# Ramanujan summation: How does 1+2+3+4... = -1/12?

**URL:** <https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903>\
**Category:** Factual Questions\
**Created:** [July 11, 2023, 4:14pm UTC](https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903 "2023-07-11T16:14:07Z")\
**Posts on this page:** 6\
**Page:** 4

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**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\
**Post date:** [July 12, 2023, 5:16pm UTC](https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903/61 "2023-07-12T17:16:17Z")

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Nice summation.

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**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\
**Post date:** [July 12, 2023, 5:52pm UTC](https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903/62 "2023-07-12T17:52:45Z")

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Thank you. Have you anything to say in addition?

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**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\
**Post date:** [July 12, 2023, 7:12pm UTC](https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903/63 "2023-07-12T19:12:48Z")

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3rd base!

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**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\
**Post date:** [July 16, 2023, 1:12am UTC](https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903/64 "2023-07-16T01:12:07Z")

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TLDR; I can get obsessive about things like this, so I investigated how do you show that \zeta(-1)=-1/12. Here is what I came up with.

I decided to try to find out why \zeta(-1)=-1/12 and came across this  
wonderful article by well-known mathematical physicist John Baez (and  
cousin of Joan, BTW):  
%[1 + 2 + 3 + 4 + ⋯ - Wikipedia](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF)  
I will try to explain how he calculates, entirely legitimately,  
\zeta(-1). To motivate it, he gives a wholly illegimate computation,  
which is actually Ramanujan’s. The reason it is illegimitate is that  
you cannot manipulate divergent series that way. You also cannot  
manipulate conditionally convergent series either. You can when the  
series of absolute values converges. As an example, you can rearrange  
the terms of 1-1/2+1/3-1/4+\cdots, which converges to \log2, to  
converge to 0 or to \pi or -\infty or anything else.

Why bother giving a bad argument? Because it motivates a perfectly  
valid argument. Watch.

Now here is what Ramanujan did. Let c=1+2+3+4+\cdots. Then  
4c=0+4+0+8+\cdots. (It is the insertion of those zeroes that would be  
legitinate in an absolutely convergent series, but not here.) Subtract  
the second from the first, to get -3c=1-2+3-4+\cdots. Now that  
series, although also divergent, can be calculated as the value at 1  
of 1-2x+3x^2-4x^3+\cdots which is  
\frac d{dx}(x-x^2+x^3-x^4+\cdots)=\frac d{dx}\frac x{1+x}=\frac 1 {(1+x)^2}  
whose value at 1 is 1/4. Since -3c=1/4, c=-1/12.

Now here is correct adaptation of that argument. Define, as in my  
earlier post, \zeta(s)=1^{-s}+2^{-s}+3^{-s}+4^{-s}+\ldots. This is  
continued, by the magic of analytic continuation to a function that  
converges–and absolutely–everywhere except at s=1. Since the  
convergence is absolute, any normal manipulation is kosher. Then  
4\zeta(s)=0+4\times 1^{-s}+0+4\times 2^{-s}+\cdots. Since the series  
converge absolutely everywhere except s=1, this insertion of zeroes is  
valid. Then subtract to get  
-3\zeta(s)=1^{-s}-2^{-s}+3^{-s}-4^{-s}+\cdots. This series is called  
\eta(s). Although this is valid for the analytic functions involved,  
the series diverge when s=-1. But it is true that  
\eta(-1)=\lim\_{x\to 1^-}(\eta(s)). This means the limit from  
the left. While  
this limit exists, the series doesn’t represent it. What to do?

At this point, Baez introduces an argument that I don’t quite follow, so  
here is my take on the rest.

One thing that you can try with a divergent series is what is called  
Cesaro summation. This actually applies to a sequence, here the  
sequence of partial sums. Basically, if a\_1,a\_2,a\_3,\ldots is a  
sequence, you replace a\_k by the average of the first k  
terms (a\_1+a\_2+\cdots a\_k)/k. This has the property that applied to  
any convergent sequence it produces a convergent sequence with the same  
limit, but applied to some divergent sequences it produces a convergent  
sequence. For example, if you apply it to 1,0,1,0,1,0,\ldots, it  
produces 1,1/2,2/3,1/2,3/5,\ldots and this converges to 1/2. This  
is one way to see that the series 1-1+1-1+1-1+\cdots can be thought to  
converge to 1/2. It is also possible that if you apply the Cesaro  
process, the new sequence still diverges, but less wildly, so that if  
you apply it again it now converges.

Let’s see how this works with 1-2+3-4+5-6+\cdots. The partial sum  
sequence is 1,-1,2,-2,3,-3,\ldots and the Cesaro sequence is  
1,0,2/3,0,3/5,0,\ldots, which obviously diverges. In fact, it is  
asymptotically 1/2,0,1/2,0,1/2,0,\ldots. It is not hard to show that  
if two sequences are asymptotically equivalent, so are their Cesaro  
sequences. If you apply the process to the latter, you get  
1/2,1/4,1/3,1/4,3/10,1/4. All the even numbered terms are 1/4,  
while the term of index 2k+1 is  
\frac{(k+1)/2}{2k+1}=\frac{k+1}{4k+2}=\frac{1+1/k}{4+2/k} which  
converges to 1/4. So the second Cesaro process converges to 1/4.

What does this do for us? Well, if you take an s\<-1 (I will restrict  
to real values of s for convenience). The series converges to  
\eta(s). If you apply the Cesaro process to it, it still converges to  
\eta(s) since the process doesn’t change the value of a convergent  
seqeunce. Apply it a second time and get another sequence that still  
converges to \eta(s). But this second sequence, no matter how  
complicated it is (and it will be a mess), converges to 1/4 when you  
put in s=-1. Since, as we saw above, -3c=\eta(1), we conclude that  
\zeta(-1)=-1/12.

Does this mean that 1+2+3+\cdots = -1/12. No, it doesn’t. It is a  
game you can play with divergent series, really signifying nothing.  
Incidentally, you might wonder what happens when you apply the Cesaro  
process to that sequence. The point is that the original sequence is  
actually convergent–to \infty --and the Cesaro process cannot change  
that for the same reason that applies to any convergent sequence.

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**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\
**Post date:** [July 16, 2023, 10:28am UTC](https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903/65 "2023-07-16T10:28:09Z")

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That “illegitimate computation”, including the Abel sum of the alternating series, does appear in Ramanujan’s first latter to Hardy.

However NB what he states in “Entry 1” is, for a sufficiently smooth function and presumably other unstated assumptions,

\sum\_{k=1}^x f(k) = c + \int\_0^x f(t)\,dt + \tfrac{1}{2}f(x) + \sum\_{k=1}^{\infty}\frac{B\_{2k}}{(2k)!}f^{(2k-1)}(x)

so the “constant” of the series should be

c = -\tfrac{1}{2}f(0) - \sum\_{k=1}^\infty \frac{B\_{2k}}{(2k)!}f^{(2k-1)}(0).

One can plug in f(t)=t, for example.

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**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\
**Post date:** [July 16, 2023, 1:00pm UTC](https://boards.straightdope.com/t/ramanujan-summation-how-does-1-2-3-4-1-12/986903/66 "2023-07-16T13:00:31Z")

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FWIW, I checked the facsimile of one of the relevant notebooks (the one where the following material is labelled Chapter VIII on page 40), and it says, modulo any transcription errors,

> [@](#):
>
> 1. f(1) + f(2) + f(3) + \cdots + f(x) = \phi(x).  
> \phi(x) = c + \int f(x) + \frac12f(x) + \frac{B\_2}{ \vert\!\underline{\,2}}f'(x) - \frac{B\_4}{\vert\!\underline{\,4}}f'''(x) + \frac{B\_6}{ \vert\!\underline{\,6}}f^{\text{v}}(x)- \frac{B\_8}{\vert\!\underline{\,8}}f^{\text{vii}}(x) + \text{&c}  
> Sol. \phi(x)-\phi(x-1) = f(x); apply VI 1.
> 
> N.B. 1. By giving any value to x, c can be found. R.H.S is not a terminating series except in some special cases. Consequently no constant can be found in \frac12f(x) + \frac{B\_2}{ \vert\!\underline{\,2}}f'(x) - \frac{B\_4}{\vert\!\underline{\,4}}f'''(x) +\text{&c} except in those special cases. If R.H.S be a terminating series, it must be some integral function of x; in this case there is no possibility of a constant (according to the ordinary sense) in \phi(x); for, \phi(1)=f(1)+\phi(0). But \phi(1)=f(1). \ \therefore \phi(0) is always 0 whether \phi(x) is rational or irrational. ∴ When \phi(x) is a rational integral function of (x) it must be divisible by x since \phi(0)=0 and consequently no constant but 0 can exist. Therefore let us try to give some other meaning for the constant of a series.
> 
> The constant of a series is the constant obtained by completing the remaining part by faithfully adhering to the above rule for summation of series. The constant of a series has some mysterious connection with the given infinite series and it is like the centre of gravity of a body. Mysterious because we may substitute it for the divergent infinite series.
> 
> Now the constant of the series 1+1+1+1+\text{&c} is -\frac12; for the sum to x terms = x = c + \int1\,dx + \frac12. \ \therefore c=-\frac12.
> 
> Another way of finding the constant is as follows— Let us take the series 1+2+3+4+5+\text{&c}. Let c be its constant. Then
> 
> \begin{align} c &= 1+2+3+4+\text{&c} \\ \therefore 4c &= \phantom{\quad+}4\phantom{+\quad}+8\quad+\text{&c}\\ \therefore -3c &= 1-2+3-4+\text{&c} = \frac{1}{(1+1)^2} = \frac14. \end{align}
> 
> \therefore c = -\frac{1}{12}.

So nothing here about analytic continuation, but he does explicitly mention finding the constant term of an asymptotic expansion; (see post #58).

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