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A lot of you will be familiar with this strange identity here: 1+2+3+
... is equal to -1/12. Now when you think about it this looks totally insane
because what we're doing here on the left side is we are adding larger and larger
positive values, infinitely many of them, so if anything this left should add up to
plus infinity. And not to negative 1/12. Most people who look at this for the
first time think that this is just nonsense. Except it comes up in a very very
famous letter in mathematics. A letter that was written by this guy down there
Srinivasa Ramanujan in 1913. It was addressed to G.H. Hardy in Cambridge, UK.
one of the most famous mathematicians at the time.
Now a lot of people have heard about this
letter but have not seen it. So I thought I show it to you. Ok, so here is the letter.
It is a long letter and it's basically lots and lots and lots of theorems, mathematical
enumeration of lots of theorems that Ramanujan says he's found and the
most remarkable thing about this is that he has no real former mathematical
training unlike Hardy. And so a lot of these things are pretty amazing and
towards the end of the letter
well this strange identity here comes up, and a couple of similarly strange
identities around it. And actually this letter got an invitation to England to
work with Hardy. So, you know, maybe we should have a really close look at this
and that's actually what I want to do today. Now, I'm not the first one to do this
Numberphile's done a couple of videos on this, and other people, but I think I'd like to
really try to do this properly, so it's going to be a very long video
and so to make it accessible to as many people as possible
I've split into four parts, so four levels of enlightenment, and you can pretty much
stop anywhere and, you know, get something out of it this way. So let me know how
far you get with this video. Since, eh, Ramanujan was a devout Hindu, I thought I'll
ask the elephant god to help us out here, to lead us through these four different
levels of enlightenment when it comes to these strange sums. Ok, here is level 1: Just do it.
Now, Ramanujan actually doesn't tell how he derives this strange value in his
letter to Hardy. You know, it's just there, he just says "that's what I get".
Now, we have his notebooks, and there is one of the pages from the notebook where
he actually talks about this. In the previous page he also talks about this; it is
more like an afterthought to what he does on the other page; but this one I
can actually do in a video like this so keep it completely elementary I can do
this one here. Let's have a closer look at what he does
Ok, so he starts by saying C. And he doesn't call it a sum he says it's a constant
C is equal to this strange sum. And then he does a couple of logical deductions, and
at the end of that he gets to -1/12. Not much, so should be able to do this
Ok, so here we go. So, he says there's something there. That something - I call it C.
C is equal to this sum. Now, if that something is halfway as reasonable,
I should be able to manipulate this as usual. One of the things I should be able to do
is multiply it by 4. So he does it. So, 4 times C is
equal to... 4 times 1 is 4. I put the 4 under the 1. No I put it under the 2. Whatever
Let's just go with it. Then 4 times 2 is 8. Again I skip one. Put the 8. Then I always skip one and put
down all the multiples of 4 here in the second line and now the next step is
to subtract the bottom from the top.
we get C minus 4C is -3C. 1-0 is 1. 2-4=-2. 3-0=3. 4-8=-4. And so on
So we get this nice sum here
1-2+3-... etc
Now comes something surprising. Now, Ramanujan
somehow sees that this is equal to a 1/(1 +1) squared. Well that's 1/4.
So, if that somehow makes sense, well, then we can say -3C = 1/4
Just solve for C, and we get -1/12
Ok, we still need that step here, that's not clear at all. So how do we get that?
That's actually has two ingredients. The first ingredient is that Ramanujan knows
that (1-2+3-... etc) sum is equal to the (1-1+1-... etc) sum squared
Ok... Second ingredient is this here: that the (1-1+1-... etc) sum is equal to 1/(1+1)
Let's try to make sense of this. First one. So, here we've got an
infinite sum that's multiplied by itself
So, how do we do this?
We have to multiply every term at the top by every term at the bottom
and then add up everything. So, let's arrange this like this
so this is the first sum here that's the second sum here. Now we do first
term times first term, that's 1 times 1 is 1
Now first term times second term is (-1)
first term times third term is 1
and so on
And that actually gives us this nice checkerboard pattern of plus and minus ones
How do you sum up this checkerboard pattern here?
Well, you do it in diagonals. So, here we go.
We're gonna put in diagonals. And now what's the sum we're actually forming here?
Well, blue diagonal is 1, green diagonal is -2, yellow diagonal is +3, next diagonal is -4
and so on. So this then clarifies what what he means by this
What about the second step?
that the (1-1+1-... etc) sum is equal to 1/(1+1)
Well, that comes from something that a lot of you will be familiar with:
this identity here
Sum of the geometric series.
Now, this is not valid for all R. It's only valid if the
R is kinda small. If it's in the range from -1 to +1, this actually works
So, for example, if you choose R is equal to 1/2, then this whole thing turns into
1+1/2+1/4+1/8+... and so on
is equal to two. So, you know, a lot of people would have seen this. A lot of people
will be familiar with this.
As I said, this works in the range from -1 to 1.
It doesn't work for -1, it doesn't work for 1. At least not in standard calculus
But what Ramanujan now does (and a couple of people before him actually)
is substitute R = -1 anyway and when you do that what you get:
Here you get -1, Here you get 1, Here you get -1 and so on
So that gives you this identity
Then, of course, if you substitute this into that, you got exactly what Ramanujan wants.
If you know any calculus, you think like every single step here is totally insane.
So there is the sums here, right. This one here, that one here, that one here
We're subtracting infinity from infinity, so nothing here makes sense.
So I've got my insaneometer here at the left.
Let's give this a score of 4 on the insaneometer.
Well, insane, yes, but
remember: this guy is a genius and this guy here
G.H.Hardy, well, he's... he's no dummy either. So somehow they can take these things
seriously, so we better have a close look.
Okay, so what does Ramanujan do here?
Well, he actually does something that a lot of mathematicians do when they don't
know what to do. So, often, you know, you're encountering expressions like this is
somebody that's kind of scribbled down on a piece of paper
something with lots of dots in it, and to start with, that's just a couple of symbols on
a piece of paper and you don't really know what they mean, right? So don't really know
what they mean. You know, if there's just finitely many (without the dots), usually know what it means
but if there's dots in there, you know, it gets tricky
You don't know what this means, right. Now, you can try and kind of just straight away
develop some theory that makes sense of these sorts of things, but often it's a
better idea just kind of do it. Okay? So what you do is you don't know what these
things are so you just let them be some sort of unknown value, and then you just
kind of go for it. So if there's anything reasonable, you should be doing this
sort of manipulations that maybe are not too wild and then maybe get you somewhere.
And let me just demonstrate this. This is kind of a very famous one. You've got this
infinite sum (that is actually a geometric series that we just looked at), and...
What can I do here?
Well, if that's anything reasonable, I should be able to multiply (just like what
Ramanujan does – at the beginning he multiplies by 4) I multiply by R
Ok? So I multiply R times C, then on the other side we get 1 times R (put R here), then R times R –
R^2 (I put it there), and so on. And that will give me this; and I do exactly what Ramanujan does
and subtract the bottom from the top. So C minus RC - there we go; then here, well,
here we are really lucky, a lot luckier than Ramanujan was with his sum. Everything
here pretty much cancels out, and the only thing that's left over is the 1.
And then we can for C, and we get this. Ok. And so, what we've found here is that assuming that
this somehow makes sense and that we can manipulate it as usual, [we] necessarily get that this has to
be equal to 1/(1-R). Now, later on in calculus, when we actually really go
for it
we actually say "well, this does make sense when R is in this range, but only
then". Yeah? So for other values of R, like -1,
that one here, that doesn't make sense; that gets discarded, that's evil, we don't want
to know this, we don't touch this in standard calculus. But actually it turns out that, you know,
if you just setup mathematics in a slightly different way, this might
actually make total sense. So on a different planet that might be totally okay.
I could also try R=2, for example, in here, and then you get something really
strange, which is very close to the sum that we actually started with; it's 1+2+2^2+
+2^3... that's all positive values getting bigger and bigger, and should
add up to infinity, but this guy here says it's actually -1. So how does that
work? Well, it turns out that if you're just on the right sort of planet, there is mathematics that makes
perfect sense and in which this is actually true.
Ok, and now, that was level 1, so you survived so far?
Perfect. Now, if you want to actually know how we make sense of all these things
stick around for Level 2 :)
Ok, so in Level 2 we're actually going to focus not so much
on what happens on other planets but what happens on Earth. So Level 2 is about following
the rules. What are the rules for dealing with these sorts of things in standard
calculus? What makes this right and all of these wrong? Well, short answer is that
this guy here on the left, which is called an infinite series, is convergent
and has sum 2, whereas all of those guys here
are divergent. Ok, let's have a close look.
Now, at some point in time, somebody wrote down this infinite
sum, and at that point in time, actually, nobody knew what it really means. Well,
you know what (1+1/2) is, you know what (1+1/2+1/4) is.
But what does it mean to add up infinitely many things? To start with,
you know, there's no... that doesn't have any meaning.
Alright, but you can try to give it to some meaning.
So we'll just start adding, ok? So we've got 1 (put it down); we add 1/2 to 1 - we get
three halves, and then we add 1/4 to get 7/4. We just keep on going like this
and we get a sequence of numbers down here that goes on forever.
And when we have a close look at this sequence of numbers, the sequence of
partial sums, then we see... Well, they get closer and closer to 2. And actually
2 is greater than all of them. And in fact, 2 is the smallest number greater than all
of these fellows, all of these partial sums.
And if you look at this expression, well, if it's to mean anything, really the most
natural thing to say here is that this should be equal to 2. And actually it's a
definition, it's, you know, a definition that mathematicians have made. It's our
choice to say that this something is equal to 2, it's a definition. It's a very
natural one, it's probably the most natural one, but it's still a choice. Our choice to do it this
way and not another way. In fact, there is another human choice in there, and
(human choices will always gonna be in the background whenever you deal with these sorts of things) it is
that we're actually always, by default, talking about real numbers.
Okay? So we're talking about real numbers; that's how we make sense of things.
There would be other possible choices for the default number system that we use, and on
different planets, maybe, they do something different; for example, the surreal numbers
or the hyperreal numbers. But for everything that we do here on earth as a default
it's the real numbers. It's also important to keep in mind.
Now, why was our choice of definition so good?
Well, because, to a large extent, you can actually manipulate these sums just like
you would manipulate finite sums. You know, this, you know, sometimes not, but...
For example, if you've got a convergent series here and its sum is A, then you
can actually multiply the right side here by something, for example 5, and you
get something convergent again. The series that we get down here is convergent again.
And it converges to... has the sum 5*A, as expected. So that's really good.
Or, if you've got two of those guys, right, so infinite series (sum A), infinite series (sum B),
we can term-wise add things, and that gives us another infinite series that converges.
And the sum is A plus B – perfect! Or we can subtract the bottom from the top
and it would be also, I think, A minus B – perfect! So what that means is
in particular, is that if this guy here is a convergent sum, then all of these
manipulations that with did here are perfectly OK, and we get the right result.
So we multiply, we subtract, it works again, and we get the right result.
Now, in standard calculus courses you actually usually don't get to this point, as actually some
bits about our definition that are not so great. So, for example, if you've got
something that works – convergent, convergent – and you multiply them
together, you would expect that you always get like A times B,
something convergent with the sum A times B, and often that is the case, but sometimes it
actually isn't. So, for example, this guy here, that's convergent and has a sum, but when
you actually do to the square of this, this sum by itself, that doesn't converge.
And it's a bit of a problem. Let's just keep this in mind
OK. So, this is maths on Earth. So let's just have a look at these sums here.
Why don't they work? Well, let's have at the first one here.
Partial sums, we need partial sums. So first partial sum is 1. 1-1 is 0, +1 is 1, -1 is 0,
and so on, so 1, 0, 1, 0, 1, 0. That doesn't converge to anything, that doesn't have a
limit. So it's, you know, it doesn't have a sum in the usual sense. So that means
it's divergent. What about the second one?
1, 1 minus 2 is -1, plus 3 is 2, minus 4 is -2, and so on. Also doesn't settle down – forget about it.
Last one. 1, 3, 6, 10 – kind of explodes.
All these series here, on the left side, are divergent. They don't have a sum.
Definitely not 1/2 here, definitely not 1/4 here, definitely not -1/12 there.
Hm, but when you kind of look at it...
These sequences that we get to, the sequence of partial sums
You know, they're very different, so instead of calculus, you could just take everything
that's not convergent and just say "well, throw it away". So you don't really look at
the difference that you get here, this, you know, graduation.
There's actually different sorts of divergence here,
different sorts of divergence. And this one is kind of a tame one, this is... Well,
Well, this is kind of oscillating around a common center here somehow.
And this kind of explodes.
You know, maybe you want to capture the differences between these
different sorts of divergence. So let's have a look at the first one here. So it's a
very very famous one; actually, the first thing you see in calculus pretty much
saying that, you know, this doesn't work, you know, for various reasons.
Now, partial sums: 1, 0, 1, 0... This is supposed to be equal to 1/2. Well, how to get one half out
of 1and 0? Well, somehow do the average.
So average of 1 and 0 is 1/2. And actually there's a very neat way
of defining the sum of one of these series in a different way. So what we do
is: we don't stop with the sequence of partial sums, but now what we do is we
generate a second sequence of averages. So, average of 1 is, well, 1; average
of 1 and 0 is one-half; average of 1, 0, and 1 is 2/3; and then 1/2 again.
And it keeps on going like this. So we have another sequence of numbers here. And
actually when you look at that one, here every second one is 1/2 and
then the red ones here
they also converge to 1/2; they have a limit of 1/2.
So overall, the sequence that we're generating here has a limit that is
1/2. So, in terms of this second sequence, what we get is convergence.
And it makes sense to say that, in some ways, this series here does converge to 1/2
And there's actually the special name for this: this sum is Cesaro convergent (if that
sort of thing happens). If you've got a series, you first do the sequence of partial sums
and then you do the sequence of averages; if the sequence of averages
converges to a number, then this number is called the Cesaro sum of the series
that we're looking at. So the series is Cesaro convergent. And actually this different
sort of convergence, just different sort of attaching a sum to a series makes a
lot of sense and really, on a different planet with different sort of
civilizations, they may actually have chosen this as their default definition of
sum of a series. Could well be. This sort of thing actually has just as nice
properties as everything else. So, for example, if a series converges in the
standard way, then it will also be Cesaro convergent. So that's really good.
But obviously what we've seen here says there are certain series that this guy here can sum
that normal, you know, normal people can't, can't sum.
Also, this sort of thing works.
If you've got something that Cesaro sums to A, then 5 times that is that one,
in terms of Cesaro.
And the adding business works, the sutracting business works. And so what that means
again is, you go back to the very beginning
If in your version of mathematics you're talking about Cesaro sums, talking about
something like this, then this calculation here will actually give you
the right answer. And it does it, right? It gives us the answer
One-half for (1-1+1...) and so on.
Further in calculus, you actually learn about these things.
They're actually very useful, so they not only make
sense but also are very useful, even for the stuff that, you know, we define as making sense.
So, for example, before I told you that when you've got something that has
sum A, and there we've got another one with sum B,
in the standard way... in the standard way. Then the product not
necessarily will converge. The product sometime diverges. But when you add the
product in the Cesaro way, you'll actually always get what you expect: A times B
So actually adding in the Cesaro way gets rid of one of those not so nice things about the
usual definition of sums. And there's other things that Cesaro summation takes care of.
There's something called Fejer's theorem, which is about
Alright, so let's have a look at our insaneometer.
Well, on planet Cesaro, if Ramanujan writes the letter, he will actually only get an
insanity score of 3, instead of an insanity score of 4 that we had before.
OK, now let's have a look at the next sum here on our list, that was the 1-2
+3, and so on, sum.
If you look at the partial sums, you get 1, -1, 2, and so on. So that doesn't work. Now,
let's just do the Cesaro thing, so the averaging. When we do the averaging, we get these guys
here, and you see: well, that doesn't work, right, that doesn't converge. So we've got every
second one as 0; these guys here, the red ones, they converge to, they have a limit of 1/2
OK, so if you actually step back here, step back and look at the sequence of partial
sums, you see every second one is a 0, and then eventually the red ones will be
indistinguishable from 1/2. OK, well, there is 1/4 that we had before
What do have 0 and 1/2 to do with 1/4? Well, it's average, right,
it's average, so we can actually kind of repeat our game. So we're just do
the averages of the averages now. So we do, you know,
put down 1, and then we do the average of those two
and then the average of those three, and then the average of those four, so this gives us another
sequence of numbers, and that will actually converge to 1/4. I'm not going to write
the numbers now, but it's going to work. And we can actually now choose these
higher-order Cesaro thing to mean our sum. Right, so on a different planet again
you know, on a different planet, maybe on the planet of the blue aliens, people
define sums of, of series in terms of this (first, second) third sequence of
numbers that's associated with any series.
Сould do that. And a planet like this
well, the insaneometer here will show just a reading of 2, so on a plant like
this, you know, that gets more and more reasonable what, what Ramanujan
does there. We can actually keep on going playing this game; so we had like the first
sequence (that's us),
second sequence (that was my green alien friend who prefers this sort of summation),
then the blue alien friend who prefers the next one down, but of course we can
do more and more kind of averages; so the next one would be the average of the average
of the average, another sequence. And these methods get more and more powerful, so you kind of
go up and up and up and up and up, and you can sum more and more of these series.
But what about this guy here? Will we ever be able to get to -1/12? Well, if you think about it,
eh, no. Right? So we've got positive numbers here, we do partial sums that's got to be
positive numbers again.
Average of positive numbers – again, positive numbers; or whatever you do here,
you'll never get anything that will get into the minuses.
Alright, for our super-sum, I have to tell you a little bit about the functions.
OK, so here we've got x^2
graphed over the positives. Now, we can extend is nice smooth function into the
negatives in infinitely many ways, and I've just drawn like three of them. So there's, we can
extend it as x^2, but also in many-many other ways, in a smooth way across 0 here.
OK? Just keep that in mind. So far, we've been talking about the real numbers. And
the real numbers are usually represented by the real number line. Now I also have to talk about the
complex numbers, which is... usually represented by the complex plane. Here
there's some complex numbers. If you've never heard of complex numbers, check out
some of these videos here before you watch the rest. Now, everything I've said
about standard summing of series, Cesaro sums, averages of averages sums
Stays true if you're actually considering series of complex numbers. You can also
have functions... complex functions defined on various bits of the complex plane. And
some of them are actually super duper nice, smooth; they're called analytic functions.
OK, so if you've got, for example, a function defined on the right part here of the
complex plane (everything to the right of the imaginary axis), how many different
ways are there to extend this to the left here, to the left part of the
complex plane? Now, these analytic functions are actually really-really-really nice,
a lot nicer than real-valued functions. So, if actually this continuation exists,
then it's uniquely determined, so that's something very-very special, and
it's called the analytic continuation of our analytic function. So these things
are super duper nice. OK, let's have a look at one of those things. So, here is
an infinite series that depends on a variable Z, so that Z can be any complex number.
Right, now we can check for which complex numbers does this thing converge, say, in a
standard way. In the standard way it converges everywhere here, in the
yellow, so, for example, at 1, Z=1, if we substitute, we get this
series, and we've already seen this before in the Apu paradox video. That sums to ln(2).
Alright. Now, since you've watched this video, you probably come up with an idea
straight away: what if we don't sum in the standard way but if we sum
like the Cesaro way? And actually something really-really nice happens – you get
part of the analytic continuation defined like this. So if you sum Cesaro, you get
everything here to the right. It's really nice. Among other things, you can figure out
what's the value of the analytic continuation here, at 0.
And what you get?
Well, at 0, we have... well, it's everywhere
1, 1, 1, 1, 1... so we get (1-1+...), and we know what the Cesaro sum of that is – that's 1/2.
OK, and then what about average the averages?
Well, that gets us everything up to here. Let's just plug in (-1) here, so -1.
Mm. Well, that sum, and so that tells us that analytic continuation here is actually equal to
1/4. And then we do averages of averages of averages, and we get this guy there... And we
keep on going like this. Actually, kind of nice homework assignment: what do these
question marks stand for? A bit hard :) Alright.
So these things are actually useful, you know, for
defining the analytic continuation of this very important Dirichlet function.
Now we'll change all the minuses to pluses; that's the Riemann zeta function.
So, the Holy Grail in mathematics – the Riemann hypothesis – is all about this guy here.
So it's very important to know what it is. Now, the series (just with a standard summation)
defines an analytic function here to the right again of the purple line.
So what does Cesaro give us? What does averages of averages give us? What do all of
the other things give us? Do they give us the analytic continuation of this thing?
Well, sadly, not really. You don't really get anywhere with this.
But there is an analytic continuation here,
and let's just see what we get you formally if we substitute (-1). So if we do
(-1), we actually get our Ramanujan sum, super sum. OK? OK, so what are we thinking now...
Well, to start with, when we defined these generalized sums, we took the
first thing that came to mind here. But what if there's different ways of
generalizing the standard way of summing series? Other ones, right, but more sophisticated ones.
Maybe those will allow us to calculate the analytic continuation of
of this sum up there. And there actually are. There are more complicated ones.
For example, Ramanujan invented one. And then... Well, since the are such sums, maybe it's
possible to then calculate these sums using
Ramanujan's that manipulation. And it's actually true: the analytic continuation
at (-1) is (-1/12).
Now here is something very-very-very important.
Those first couple of summations methods
that I've talked about... They conceivably could really be replacements of the
standard way of summing things (on some planet). Now, these more complicated
summation methods, like Ramanujan's one (when you look at it: it's got integrals, and it's got
like sums in it, and special numbers), they only work in special context. They work for
special sorts of series. They're really-really useful there, but they could never ever
replace standard summation. If at any point in time you get on your calculus
test "what is 1+2+3+4 etc.", and you answer "-1/12", almost certainly
you'll get zero marks. Whenever you get something like this, there's a standard way of
interpreting a sum like this; and the standard way involves, you know, the
partial sums. Do the partial sums converge or not over the reals – that's
what's being asked here. Nothing else, OK? If you give any other answer, it will be wrong.
OK? Let's keep that in mind.
So the last thing I want to talk about is again the
geometric series because most of you will be familiar with this, and most of
you have been told to disregard anything outside, you know, small values of Z.
Now, this one actually works for complex numbers too and defines a nice analytic
function here in the unit circle. If you actually now some using Cesaro, it will
also give you things on the unit circle. So it will actually give you the
analytic continuation on the unit circle. What else? Well, if you do
averages of averages, actually doesn't get you anymore.
But there's an analytic continuation, of course, of this function that's defined here.
And this analytic continuation is actually this sum here that we calculated before.
So again, I mean, what comes out of there is not nonsense; it defines something, it
defines values of the analytic continuation of the function that's defined
by this guy here. But that's what it is.