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Welcome to another Mathologer video today I'd\h like to tell you about a bunch of very paradoxical\h\h
properties of the so-called harmonic series. This\h monster is one of the most iconic infinite objects\h\h
in mathematics and has been investigated\h for hundreds if not thousands of years.\h\h
Of course everybody who knows some calculus\h will be familiar with some basic properties\h\h
of the harmonic series. However most of its\h really amazing paradoxical properties even\h\h
many calculus professors don't know.\h\h
The reason why I'm telling you about the harmonics\h series today are some more counter-intuitive facts\h\h
that were only discovered recently and that I\h only found out about two months ago. This video\h\h
has six chapters each with its own highlights.\h Let's have a vote at the very end to figure out\h\h
which of these highlights you like best.\h Will be interesting and by participating\h\h
you'll automatically enter into a draw for\h a chance to win a copy of one of my books.\h\h
As a bit of a warm-up let's engage in a little\h thought experiment. Over there i’ve put weights\h\h
of one and two kilos on a simple balance.\h Let's assume that the gray bar does not\h\h
weigh anything. Remember thought experiment we\h can do this :) if i remove the supporting hands\h\h
will things stay balanced? Well obviously not. The\h red two is heavier so it will go down like that.\h\h
So to balance things i have to move\h the pointy fulcrum to the right.\h\h
But where exactly do i have to put it. Now if\h you're familiar with archimedes law of the lever\h\h
you shouldn't have any trouble figuring\h this out for yourself. Anyway the answer\h\h
is one-third from the right. Now here's a nice\h two-glance way to see that this is really true.\h\h
Put the 2 kilo weight on a little tray like\h this. Again we'll assume that the tray does not\h\h
weigh anything. Thought experiment remember. Now\h split the 2 kilo weight into two 1 kilo weights.\h\h
Keep moving them apart up to here. Obviously\h the combined center of mass of the two one kilo\h\h
weights is still where it used to be. There right\h in the middle. On the other hand all the three one\h\h
kilo weights are now equally spaced. Can you see\h it? There the spacing on the left is two thirds\h\h
and the spacing on the right is one third plus\h another one third mirrored over is two-thirds.\h\h
But then the combined center of mass of the three\h equally spaced one kilo weights is right in the\h\h
middle that is exactly at the one-third mark from\h the right. A very very pretty two glance proof\h\h
don't you think? And the whole thing generalizes\h nicely? If instead of a two kilo weight we have a\h\h
three kilo weight on the right then the balancing\h point is one fourth from the right. There\h\h
put the tray, spread them out, equally spaced,\h right in the middle, works. I like this stuff.\h\h
And with a four kilo weight on the right\h the balancing point is at one fifth.\h\h
Okay when i teach the harmonic series at uni i\h usually motivate it with a nice real world puzzle,\h\h
versions of which you may encounter in many\h science museums the brown blob over there\h\h
is supposed to be the edge of a cliff. Now add\h 20 equal rectangular blocks to this scenario.\h\h
What we want to do is to rearrange the blocks into\h a stack with as large an overhang as possible.\h\h
What do i mean by this? Well i could simply\h push the whole stack to the right like this.\h\h
This will create this much of an overhang.\h There the green bit that's the overhang. Can\h\h
we do better? Yes for example we can push\h out the one block on top just a little.\h\h
There now the green overhang is larger.\h Of course we've got to be careful.\h\h
Push things over too far and the stack\h will topple over. And since the blocks\h\h
are not connected to each other the toppling can\h happen at any level. For example right up there.\h\h
So what's the furthest overhang we can create?\h I know hard to tell given only what I've told\h\h
you so far. Well so as usual let's look at some\h small stacks first to build up some intuition\h\h
you know the drill. Right from last time. Now\h there a stack with just one block. How far can\h\h
we push this block to the right? Easy the center\h of mass of this block is right in the middle\h\h
and the block will be stable as long as the center\h of mass is above the cliff. This means we can push\h\h
the block all the way up to here. Now let's say\h our blocks are two units long. Then our maximal\h\h
overhang is half that. So one unit overhang. Okay\h what about stacks consisting of two blocks? Well\h\h
with two blocks we can definitely also get one\h unit overhang. Right just slide the second block\h\h
in like this. There that obviously works, a two\h block stack with an overhang of one. But we can\h\h
definitely do better because the center of mass of\h the whole two block stack in front of us is right\h\h
here in the middle. Well because the whole thing\h is symmetrical it has to be in the middle right.\h\h
And this means that we can slide the whole thing\h further over to the right up to this point. Right\h\h
the center of mass of the whole stack is above\h the cliff edge and so nothing will fall down.\h\h
What's the overhang now? Well the one from\h before plus one half. And now we just repeat\h\h
to get larger and larger overhangs. So\h first add a third block in like this.\h\h
Next find the center of mass of this new\h three block stack. Now unlike in the cases\h\h
of just one and two blocks where everything\h was nice symmetrical it's not clear where\h\h
the center of mass is. However using our warm-up\h exercise we can easily figure out its location.\h\h
So we can think of our three block\h stack as consisting of two parts.\h\h
The new block highlighted in blue and the old\h two blocks stack. Now the center of mass of the\h\h
two blocks stack is already part of the picture.\h The center of mass of the blue one is here right\h\h
in the middle. How far apart are the centers? Well\h that's half the length of the blue block so that's\h\h
one. Next our two stack weighs twice as much as\h the blue block and this means we're dealing with\h\h
exactly our warm-up setup. And this means that the\h combined center of mass of the three block stack\h\h
is one-third in from the cliff\h edge. Ah there nice animation\h\h
nice. Now push to the right like this and\h we get an overhang of one plus one half plus\h\h
one third. Repeat again there slide in, find\h the center of mass, nice animation again, and\h\h
yes slide over and at this point the overhang is\h one plus one half plus one third plus one fourth\h\h
and we now know exactly how this will continue.\h However also notice that at this point\h\h
the top block is completely beyond the cliff edge.\h\h
And this is pretty amazing isn't it? A challenge\h for you: actually perform this experiment. So use\h\h
identical brick-like objects like books to build\h a stack at the edge of a table in which the top\h\h
block completely clears the table. Link to a photo\h of your little stacking miracle in the comments.\h\h
Pretty tough challenge. Anyway rinse and repeat\h and we get this remarkable leaning tower.\h\h
And the overhang we can achieve this way with\h 20 blocks is? Well one plus one half plus one\h\h
third all the way to one twentieth which is\h approximately three point six. Very cool.\h\h
What I've just shown you is a Mathologerization\h of a construction that's at least\h\h
170 years old. These days the stack over there is\h often referred to as the leaning tower of lire.\h\h
This name was coined by the physicist Paul\h Johnson and was the title of a note by him\h\h
that appeared in the American Journal of\h Physics in 1955. There that's the article\h\h
the lire in the title was the name of the Italian\h currency prior to introduction of the euro and in\h\h
his paper Johnson actually plays with a stack\h of coins rather than a stack of bricks. Anyway\h\h
now a very natural question to ask is how much of\h an overhang can we produce if we have an unlimited\h\h
supply of blocks. To answer this question we\h have to evaluate the sum of the harmonic series.\h\h
There she is, the lead character in\h today's tale of mathematical mystery.\h\h
I'll puzzle out the sum together with you in a\h moment but not before i say a little more about\h\h
our original puzzle. To find a maximal overhang\h by stacking these 20 blocks over there. I'm fairly\h\h
sure that 99% of all the people who are familiar\h with this leaning tower are under the impression\h\h
that this tower produces the largest possible\h overhang using 20 blocks. This is not the case,\h\h
not even close. Want to see what the optimal\h stack really looks like? Ready to be disgusted?\h\h
Here's the optimal stack which was only\h discovered something like 10 years ago.\h\h
The same 20 blocks but reaches out\h much further. And ugly as hell right?\h\h
And didn't they always tell you that all\h optimal maths is automatically beautiful?\h\h
While you often encounter beauty\h in the optimal this is definitely\h\h
not always the case? But of course beauty is in\h the eye of the beholder and there's definitely a\h\h
lot of ingenuity to be admired here. For example\h the use of counterweights and bridging weights\h\h
to weigh down pieces that would otherwise\h fall down. What the presence of these\h\h
bridging pieces also implies is that you cannot\h actually build this weird stack layer by layer\h\h
from the ground up. Right, building up to here\h the piece in front would definitely fall down.\h\h
Now whenever i talk about this someone will ask\h whether it is at least true that the leaning tower\h\h
is the optimal stack if we allow only one brick\h per layer. And this is actually true. Although a\h\h
correct proof that this is true for all possible\h numbers of bricks was only nailed down in 2018\h\h
by David Treeby somebody from Monash, in an\h article in the American Mathematical Monthly.\h\h
That this maximal overhang property is not\h completely obvious is hinted at by the fact\h\h
that there is actually another one block per\h layer stack that has exactly the same overhang\h\h
as our leaning tower of lire. Again hardly anybody\h seems to be aware of this. Have another look at\h\h
the two top blocks. Can you see where I'm going\h with this? Center of mass right in the middle.\h\h
Well simply flip them like this and\h you get that second maximal stack.\h\h
Cute right? If you're interested in\h chasing down this particular rabbit\h\h
hole I've linked to some nice links\h in the description to get you started.\h\h
So we can arrange for one complete brick to\h be beyond the cliff edge using four bricks.\h\h
How can you do the same with only three\h bricks? Give you solutions in the comments.\h\h
Okay so we would like to know how much of an\h overhang one of our towers of lire can have if\h\h
we have an unlimited supply of blocks. To answer\h this question we have to figure out what the sum\h\h
of the harmonic series is. Well let's start\h adding. Okay so we are adding more and more\h\h
positive terms and so obviously these partial sums\h will get bigger and bigger. This means there are\h\h
really only two possibilities. One, these partial\h sums converge to a finite number which we would\h\h
then declare to be the sum of the series. Or\h the partial sums explode to positive infinity.\h\h
Which one is it? Obviously many of you will know\h the answer but imagine for the moment that this is\h\h
the first time you see the series. What would\h you bet your life on? Finite or infinite sum?\h\h
Well the terms we're adding get smaller and\h smaller and amount to closer and closer to\h\h
nothing, so shouldn't the partial sum settle\h down to something finite, just like for some\h\h
other famous subseries of the harmonic series,\h like this one here. What's the sum of this one?\h\h
Well pretty obvious… so there, infinitely many\h positive terms adding up to something finite.\h\h
What about the sum we're interested in.\h Well let's not drag it out any further.\h\h
Turns out the harmonic series explodes to\h infinity and the first in my opinion nicest\h\h
proof of this fact goes back almost 700 years\h and is due to the french bishop Nicole Oresme.\h\h
Hmm clearly some posture problems due\h to excessive obsessing with mathematics.\h\h
Okay here's this proof animated\h as best as i can. Enjoy.\h\h
An absolute gem of a proof\h don't you think? Of course\h\h
what this implies is the leaning tower\h paradox, that at least in theory we\h\h
can make our overhang as large as we\h wish by adding more and more blocks.\h\h
Marvelous stuff isn't it? There are\h actually a couple of other spectacular\h\h
real world paradoxes that are based on the\h fact that the harmonic series is exploding.\h\h
For example if you have not heard\h of the worm on a rubber band paradox\h\h
definitely check out the links in the description\h of this video. And if you know of any off the\h\h
beaten track real world harmonic paradox please\h share it with the rest of us in the comments. Okay\h\h
what have we got so far? One glance balancing,\h leaning tower of lire, bizarre maximal overhang\h\h
towers, and a 700 year old proof by a\h bishop. What do you like best so far?\h\h
Okay so the harmonic series diverges to infinity,\h but it does so extremely slowly. Have a look.\h\h
As you can see the series really grows very\h slowly. That's a million terms we've added here.\h\h
How many terms do you think it takes\h for the partial sum to hit 100.\h\h
That number was calculated for the first\h time in 1968 by the mathematician John w\h\h
Wrench jr which was a bit of an achievement\h at the time. Here it is. So to get to 100\h\h
you have to add in the order of 10 to the 43\h terms. That's really mind-boggling isn't it?\h\h
Also how did John Wrench figure out that\h exact number? Obviously actually adding\h\h
the first 10 to the 43 terms one at a time is\h completely out of the question. Even the most\h\h
powerful computers would take gazillions of\h years to add up that many terms one by one.\h\h
Well maybe there's some kind of magical formula\h for the n's partial sum that allows us to solve\h\h
for the answer? Such a formula would also allow\h us to figure out that all important question how\h\h
much of an overhang a leaning tower of one google\h blocks that is 10 to the 100 blocks will produce.\h\h
Well maybe not that important\h question itself but formulas like this\h\h
really are incredibly important. Anyway\h let's hunt for such a magical formula.\h\h
If you watched the last what comes\h next video you know what comes next.\h\h
We start summing again from the beginning but\h this time write down the partial sums as fractions\h\h
let's see whether we can guess a\h pattern that translates into a formula.\h\h
Okay here comes the dreaded question: can you see\h a pattern? Well looks like a complete mess doesn't\h\h
it? But there's something striking about all these\h fractions. Can you see it? No okay here's a hint.\h\h
What do all the numerators have in common and\h what do all the denominators have in common?\h\h
Yeah of course many of you will have spotted\h straight away that all the numerators are odd\h\h
and all the denominators are even. Coincidence?\h I don't think so! In fact if you keep on going\h\h
you keep getting partial sums that\h are odd divided by even fractions.\h\h
Interesting but so what? Well actually in terms\h of odd and even there are four different types\h\h
of fractions. Odd over even, even over\h odd, even over even and odd over odd.\h\h
There's one thing that is special about odd\h over even fractions and that is … well, think\h\h
integers … if you divide an odd number by an even\h number the result can never be an integer. Right?\h\h
Whereas with the other types of fractions\h they can actually be integers in disguise.\h\h
There all those fractions are really integers\h in disguise. Nice little insight isn't it?\h\h
An odd divided by an even\h fraction cannot be an integer.\h\h
That could be a starting point for a nice\h number magic trick. Have to ponder this.\h\h
Anyway once you notice this odd over even pattern\h in the sequence of partial sums it's actually\h\h
not hard to prove rigorously that in fact all\h infinitely many partial sums are of this form,\h\h
with the exception of the first partial\h sum 1 of course which is an integer. And\h\h
this then also shows that apart from that\h 1 at the beginning there are no integers\h\h
among the partial sums. That's a pretty\h amazing result when you think about it.\h\h
Right? So these partial sums creep up slower\h and slower smaller in smaller increments.\h\h
On the way to infinity they pass every\h single one of the infinitely many integers\h\h
and manage to miss them all by my minuter and\h minuter amounts. Amazing isn't it? But in fact\h\h
something even more remarkable is true. Add up a\h randomly picked couple of consecutive terms of the\h\h
harmonic series like these ones here. Consecutive\h is the important bit here. Then it can be shown\h\h
that all of these sums combine into odd over\h even fractions and are therefore never integers.\h\h
The first proof of this fact was published in\h 1918 and is due to the Hungarian mathematician\h\h
Joseph Kurschak the very friendly looking guy\h on the white chair in the picture. That's a very\h\h
typical mathematician group photo 100 years ago:\h all men, ties and suits and no t-shirts. Times\h\h
have definitely changed. Okay to finish off this\h chapter here is another challenge question for\h\h
you: can you write the numbers 2 and 3 as sum of\h reciprocals of distinct positive integers. Hmm i\h\h
guess i got a bit sidetracked there. Still i think\h this detour was totally worth it and this is also\h\h
a nice example of how scientific discovery works.\h You don't always find what you're looking for.\h\h
Anyway remember what we set out to find at the\h beginning of the last chapter was an easy formula\h\h
for the n's partial sum of the harmonic series.\h The usual approach to look at a couple of small\h\h
examples to try and spot a pattern that amounts to\h a formula actually does not work for the harmonic\h\h
series but there is another very nice and natural\h way to approach this problem. One, one half, one\h\h
third, one fourth 1/n s. Doesn't this cry out for\h us to ponder our sum in terms of the function 1/x.\h\h
Well let's plot it. Where are the terms of our\h sum hiding in this picture? Well plug in one,\h\h
one over one is one so there is one.\h Obviously that blue segment is one unit long\h\h
but there's another one hiding here, a one unit\h area. Can you spot it? There that's square.\h\h
One half is here, and as an area here.\h One half times one is one half and so on.\h\h
And so there that's a simple visual area\h counterpart to the harmonic series. Of course\h\h
there is another area in this picture that cries\h out to be highlighted that area under the curve.\h\h
But how is that going to help? Well we're\h chasing the partial sums of the harmonic series.\h\h
Let's focus on a small example. 1 plus 1\h half plus one-third plus one-fourth is equal\h\h
to the blue area. Turns out that there is a really\h easy formula for the corresponding area under the\h\h
curve. Integral baby calculus 101 a lot of you\h would have done this in school tells us that this\h\h
area is exactly equal to the natural logarithm of\h 5. And of course that means that our partial sum\h\h
is approximately equal to that logarithm. In\h general this approximation looks like this.\h\h
How much of a difference is there between the left\h and right sides? Well that difference is the area\h\h
of the remaining blue bits sticking out on top\h and luckily these blue bits don't add up too much.\h\h
Let me show you. See that one times one square\h over there? Well all the blue bits fit in there\h\h
and so the sum of the blue bits, the difference\h up there is less than one. And that's important no\h\h
matter how humongous the partial sum is. In fact\h this tethering of the logarithm and the partial\h\h
sum immediately implies another proof of the fact\h that the harmonic series explodes to infinity.\h\h
Right, since the logarithm explodes to infinity\h the tethered partial sum gets dragged along\h\h
to infinity doesn't have a choice at all in\h this matter. Anyway since for large partial\h\h
sums an extra one doesn't really matter, for\h all intents and purposes the n's partial sum\h\h
is equal to the natural logarithm of n plus one.\h Having said that we can still improve this simple\h\h
very good approximate formula\h dramatically with very little effort.\h\h
Notice that really all infinitely many blue bits\h fit into the square on the left. This means that\h\h
all the blue bits add up to a number less than one\h this number has a special name it's called gamma,\h\h
or the Euler-Mascheroni constant or simply Euler's\h constant. Euler again can't escape him right.\h\h
Another challenge for you: okay so gamma is less\h than one. That’s obvious right. But can you also\h\h
see at a glance that gamma has to be greater than\h 0.5. If you can, tell us why this is obvious.\h\h
Anyway Euler was the first to investigate\h gamma around 1734 and calculated the first five\h\h
decimals. Gamma is one of those mathematical super\h constant that just like pi, phi and e pops up all\h\h
over the place. Just as a little teaser here is\h one of my favourite identities involving gamma\h\h
they are e gamma and all the prime\h numbers in one formula. How crazy is that?\h\h
Anyway here is a super nice trick let's add\h gamma to our approximate formula for the\h\h
partial sums of the harmonic series. Here's what\h the sum corresponds to in the picture so the sum\h\h
is exactly equal to the pink partial sum one\h plus one half plus one third plus one fourth,\h\h
plus a tiny portion of gamma corresponding to the\h extra blue bits petering off to the right.when\h\h
you think about this this is really an amazing\h formula. As n and with it the partial sum gets\h\h
huger and huger this formula approximates\h the partial sum better and better. That's\h\h
the exact opposite to what most formulas\h approximating exploding sums usually do.\h\h
Just two examples to show you how amazing this\h formula really is for n is equal to 5 we get this.\h\h
So in terms of correct digits the formula\h gets the integer part of the partial sum\h\h
right but no decimals. Now here's what you\h get for n is equal to five hundred thousand.\h\h
That's already five correct decimals and it\h really gets better and better the larger n is.\h\h
Absolutely marvellous isn't it? To finish this\h chapter let me tackle those two questions that we\h\h
set out to answer using our formula. Remember the\h humongous exact number of harmonic terms you have\h\h
to add to get a sum exceeding 100?\h That number John Wrench calculated.\h\h
What does our formula suggest this number is? Well\h we want 100. Let's solve for it. There autopilot…\h\h
and well if you actually evaluate the expression\h on the right for example in Mathematica you arrive\h\h
at a number that's just shy of John Wrench's exact\h number. According to Mathematica the difference\h\h
is about 0.63. Pretty good. Our gamma is really a\h very magical constant. In fact John Wrench's proof\h\h
that his number is it is based on our formula,\h starts with our formula. Of course we can also use\h\h
our formula to figure out how much of an overhang\h a leaning tower with n blocks can produce.\h\h
So how about one google blocks for example. Well\h the natural logarithm of 10 to the 100 is 230\h\h
and a bit units this thing really grows very\h slowly. Well that's logarithmic growth for you.\h\h
In comparison it has been shown that\h there exists a positive number c\h\h
such that by using crazy towers an overhang of c\h times cube root of n is achievable for large n.\h\h
That's exponentially better than logarithmic\h growth. Nobody knows exactly what the absolute\h\h
best crazy tower looks like for large numbers of\h blocks but one not so crazy family whose overhang\h\h
grows like cube root of n are these parabolic\h towers that were discovered by mike paterson\h\h
and Uri Zwick around 2009. The example over there\h consists of 571 blocks and produces an overhang of\h\h
10 units the smallest leaning tower that has\h this overhang contains 12 367 blocks. Good stuff\h\h
and there are still lots of nice open problems\h that need to be sorted out. Maybe by you. So\h\h
to finish this chapter let me at least mention\h one more super interesting fact about gamma.\h\h
Since you are watching this you're\h probably all aware of the fact that there\h\h
is a mysterious and important connection\h between the sum of the positive integers\h\h
one plus two plus three and the number minus\h one-twelfths. Despite what some people claim\h\h
this wonderful relationship is not an equality.\h Now when we dig deeper we find that in a very\h\h
precise sense -1/12 is to the sum of the positive\h integers what gamma is to the harmonic series,\h\h
the sum of the reciprocals of the positive\h integers. I don't have any time to elaborate\h\h
on this today. If you'd like to do a bit of\h research on your own google Ramanujan summation\h\h
and find out why the Ramanujan sum of one plus\h two plus three -1/12 and the Ramanujan sum of the\h\h
harmonic series is gamma. Okay what have we got so\h far: one glance balancing leaning, tower of lire,\h\h
bizarre maximal overhang towers, a 700 year old\h proof by a bishop, no integers among the partial\h\h
sums, magical approximate formula and of course\h gamma. And i even managed to squeeze -1/12s in.\h\h
What do you like best so far? Remember\h there will be a vote at the end.\h\h
Okay final chapter: to finish off let me just tell\h you about that other remarkable harmonic fact that\h\h
i only found out about two months ago and that\h inspired me to dedicate this video to my good\h\h
old friend the harmonic series. The harmonic\h series is the sum of sums in the sense that\h\h
it contains many many many many very important\h infinite series: for example there's the geometric\h\h
series of the powers of one half there which sums\h the two. Then there's Euler's super famous sum of\h\h
the reciprocals of the squares, the reciprocals\h of the fourth powers and in general all those\h\h
other reciprocals of integer powers series\h that are part of the Riemann zeta function.\h\h
Then there is e as the sum of the reciprocals of\h the factorials and so on and so forth. I've talked\h\h
about these and many other famous subseries\h of the harmonic series in previous videos.\h\h
Now obviously there are infinitely many sub-series\h contained in the harmonic series summing to\h\h
infinitely many numbers and so a natural question\h to ask is whether every number is such a sum.\h\h
What do you think? What's your gut feeling? The\h answer is yes every positive number occurs as a\h\h
sum of infinitely many sub-series of the harmonic\h series. This is actually very very easy to see?\h\h
Let me quickly show you how you can find a\h sub-series that sums to our new favourite magic\h\h
number gamma using a greedy algorithm. Works for\h any other positive number. As the first term of\h\h
a subseries that will sum to gamma choose the\h largest harmonic reciprocal less than gamma.\h\h
Well gamma is 0.57 and so the largest\h harmonic reciprocal less than gamma is\h\h
one-half. The next term of our\h sub-series is the largest harmonic term\h\h
that when added to one-half gives the sum\h less than gamma. That happens to be 1/13.\h\h
There that sum is just a little bit less than\h gamma and you just keep on going like this, always\h\h
add the largest possible harmonic term that gives\h you a sum less than gamma. This will automatically\h\h
generate a subseries containing infinitely\h many terms that adds very rapidly to gamma.nice\h\h
and super simple this is an example of a\h greedy algorithm in action. We always greedily\h\h
go for the largest possible reciprocal\h that fits. Of course apart from all those\h\h
sub-series with finite sums there are also plenty\h of sub-series that just like the harmonic series\h\h
diverge to infinity. For example the sum of\h the reciprocals of the even numbers diverges.\h\h
Pretty obvious because the series is also just\h the original series, there's the original series\h\h
again, multiplied by one half. So the sum of the\h reciprocals of the even numbers is one half times\h\h
infinity which is still infinity. What about the\h sum of the reciprocals of the odd numbers? Well\h\h
the first odd reciprocal one is greater\h than the first even reciprocal of one half.\h\h
The second odd reciprocal one third is greater\h than the second even reciprocal 1/4 and so on.\h\h
And so the sum of the odd reciprocals is\h greater term by term than the infinite sum\h\h
of the even reciprocals and so also has to be\h infinite. Okay maybe that's not that terribly\h\h
surprising. Here's something trickier. How\h about the sum of the reciprocals of the primes?\h\h
Finite or infinite? Well what do you think? Of\h course there are infinitely many primes but they\h\h
are pretty sparse with arbitrarily large gaps in\h between them. Does this mean the sum is finite?\h\h
Well those of you who watched my video on Euler's\h product formula for the Riemann zeta function\h\h
may remember that we proved,\h following Euler's footsteps of course,\h\h
that the surprising answer is this sum is\h infinite. Really not obvious at all. Let's\h\h
see how much of an intuition you've developed\h for these infinite sums. I'll show you a couple\h\h
of sub-series and you try to guess whether\h they have a finite or infinite sum. Ready?\h\h
Okay that's the full harmonic series over\h there. First let's skip every tenth term\h\h
starting with one ninth. That takes\h out this infinite column of terms.\h\h
Is the leftover series finite or infinite?\h You got it? Easy right? This sub series must\h\h
explode because, well, there are lots of\h ways to see this at a glance. For example\h\h
it must explode because it contains all the even\h reciprocals which we've already seen do explode.\h\h
For my second mystery series i’m going\h to throw away all the reciprocals of\h\h
numbers that contain the digit nine. Okay\h so that gets rid of the last column again.\h\h
Right all those have a nine. But of course there\h are numbers other than these that feature the\h\h
digit nine. For example all the guys at the bottom\h there. Okay no more nines as digits left over.\h\h
Finite or infinite? Imagine your life depends on\h getting this right. You've got five seconds five\h\h
four three two one zero. You got it? Well based on\h past experience asking this question almost all of\h\h
you will have gone for infinite. Which is wrong.\h Believe it or not but this one has a finite sum.\h\h
I first encountered this no 9th sum in the\h second week at uni in Germany a couple of\h\h
ice ages ago. They actually gave it to us as\h a homework exercise. This fun series is known\h\h
as Kempner’s series named after the mathematician\h Kempner who published a finiteness proof in 1914.\h\h
What makes this counter-intuitive results\h particularly memorable for me is that i\h\h
did come with a finiteness proof myself\h as part of this homework assignment.\h\h
I can't remember how i proved it but i do remember\h that i got zero marks because the marker could not\h\h
be bothered to actually engage with my proof that\h was different from what was expected. Grrrr :(\h\h
I'll do a super nice animated algebra proof\h that the no 9 series has a finite sum at the\h\h
end of this video but just real quick one way\h to intuitively see that Kempner's no 9 series\h\h
may be sparse enough to allow for it to converge\h is to realize that the picture over there is a\h\h
bit misleading, that in fact the vast majority\h of integers with large numbers of digits will\h\h
contain a nine among their digits and therefore\h the reciprocals will not count towards our sum.\h\h
Tristan one of the three volunteers who helped\h me with proofreading and polishing my slideshows\h\h
suggested a nice visual way to illustrate\h this thinning out. Make the 10 x 10 picture\h\h
over there a bit more squarish and color the\h part with the remaining reciprocals black.\h\h
Let's pretend the upper left square is also\h black. Makes for a cleaner argument. Okay\h\h
then up to that little fudging in the corner\h the black area inside the big square tells\h\h
us how many of the first 10 times 10 terms of\h the harmonic series are contained in Kempner's\h\h
no 9s series. Zooming out the corresponding\h picture for the first 100 times 100 that's the\h\h
first 10000 terms of the harmonic series looks\h like. There is more white than before and so\h\h
the percentage of omitted terms has gone up.\h Interesting. Here's the picture for one thousand\h\h
times one thousand that's the first one million\h terms. Of the harmonic series the percentage of\h\h
omitted terms has gone up again, now ten thousand\h times ten thousand. Looks the same but it's not.\h\h
On closer inspection there are actually tiny\h white subdivisions of the little black squares\h\h
that mirror the nine times nine subdivision\h that you're seeing in front of you. Like for\h\h
example the little square right there. When you\h have a close look it actually looks like this.\h\h
Because of the presence of the white subdivisions\h the little square is actually gray and not\h\h
completely black and so again the percentage of\h white has gone up. As we zoom in digit by digit\h\h
the subdivisioning continues turning what you\h see in front of you into a fractal-like object.\h\h
And the percentage of white will go to\h 100 percent as we push things to infinity.\h\h
Pretty amazing. Thanks again to Tristan for\h suggesting this fun visualization of the really\h\h
dramatic thinning out of the harmonic series when\h we throw away all the reciprocals of integers that\h\h
contain a digit nine. Makes the convergence of\h the series so much more believable. Not a proof\h\h
but as i said I'll do an animated algebra proof at\h the end. Okay everything I've told you so far I've\h\h
known for many many years now. What i stumbled\h across two months ago were a couple of articles\h\h
by the mathematician and computer scientist\h Robert Baillie full of very interesting and\h\h
surprising results about the no 9 series and a\h whole universe of closely related crazy sub-series\h\h
of the harmonic series that i’d never heard of\h before. For example it turns out that nobody\h\h
actually knew the sum of the no 9 series until\h quite recently. What Kempner showed in his 1914\h\h
article was that the sum of the no 9 series is\h less than 90. How close is 90 to the actual sum.\h\h
Well although the no 9 series turned into being\h a bit of a celebrity among people in the know,\h\h
nobody appears to have given this question\h any serious thought, until Robert Baillie\h\h
did so about 30 years ago. He figured out that\h the sum of the no 9 series is approximately\h\h
22.92. At that point in time he also figured out\h good approximations of the no zeros, no ones, no\h\h
twos etc series. Here's a table from his article\h which summarizes his results. Why had nobody\h\h
calculated any reasonable approximations before?\h Well it turns out that all these missing digits\h\h
series converge at a snail's pace. For example for\h the no 9 series the sum of the first 10 to the 28\h\h
terms is still less than 22, so not even close\h to what most of us would call a reasonable\h\h
approximation of the true value of the sum.\h And so for these series it's also completely\h\h
unfeasible to simply keep adding up terms until\h we get close to the true value. You really need\h\h
a good idea to get anywhere with approximating\h these sums. More recently Robert Baillie also\h\h
managed to calculate similarly good approximations\h for other closely related crazy series like the\h\h
no zeros and no nine series (so a couple of digits\h missing not just one), the at most 10 9s series\h\h
(so every term has at most 10 nines), the exactly\h 100 0s series, and so on. Really impressive stuff\h\h
especially since most of these series converge\h much much much slower than the already very\h\h
slowly converging no 9 series. For example the\h sum of the 100 0s series is approximately 23.03.\h\h
Also pretty amazing in another way when you think\h about it right? The first term of the 100 series\h\h
is one over one google which of course is a\h crazy small number and yet the sum of the series\h\h
in which every term is exactly 100 zeros ends up\h being larger than the sum of the no 9 series which\h\h
starts out with much larger terms. So how are\h these really good approximation calculated? Well\h\h
it's not that terribly complicated but\h also quite nitty gritty technical so i\h\h
won't go into details here. The basic idea\h is to exploit the patterns in these series\h\h
in the calculations, the kind of patterns that\h were visible in Tristan's fractal for example.\h\h
For the keen among you I've included some\h relevant links in the description of this video.\h\h
Anyway as you can probably tell Robert Baillie's\h papers really made my day when i stumble across\h\h
them and i hope I've been able to convey to\h you some of my excitement about these new\h\h
and old insights into the nature of the harmonic\h series and infinite series in general. What i’ll\h\h
finish with today is a Mathologerized slightly\h refined version of Kempner's original proof\h\h
which shows that the son, sum not the son\h of the no 9 series is less than 80. Watch\h\h
it and afterwards cast your vote for the harmonic\h marvel in this video that you enjoyed most. What's\h\h
most memorable: one glance balancing, leaning\h tower of lire, bizarre maximal overhang towers,\h\h
that 700 year old proof, no integers among the\h partial sums, amazing approximate formula, gamma,\h\h
greedy algorithm, Tristan's fractal, finiteness\h proof. Well we really did cover a lot of ground.\h\h
Cast your vote by leaving a comment within\h the first two weeks after this video\h\h
went live for a chance to win one of\h my books. I'll then pick a random name\h\h
among those who participated and a random\h book and will I’ll announce the winner\h\h
as part of the 2020 Christmas video. Okay now to\h finish here's why Kempner’s no 9e sum is finite.