Evaluating, $\sum_{n=1}^{\infty} n$

Question

Evaluating:

$$S = \sum_{n=1}^{\infty} n$$

Apparently, $S = \zeta(-1) = -\frac{1}{12}$

Which is crazy. How can $1 + 2 < 1$ ?

Anyway,

How do you evaluate $S = \zeta(-1)$

Hint please? Thanks

That one is a classic, due to Euler. As @CameronBuie says, it must have been asked lots of times. My searching skills aren't up to actually finding the duplicate questions, though … — Harald Hanche-Olsen, Feb 22 '15 at 14:41
@HaraldHanche-Olsen, how does it work? I clearly dont see how $1 + 2 + 3... < 1$? — Lebes, Feb 22 '15 at 14:42

score 3 · Answer 1 · answered Feb 22 '15 at 14:40

$\zeta(-1)$ is not what you think if is, that is, is $\not= \sum_{n\geq 1} n$. So that there is NO contradiction. You should rather look for how to extend holomorphically $\zeta$... Plenty of stuff about that on the internet... See this :

http://mat.uab.cat/matmat/PDFv2009/v2009n06.pdf

for instance.

score 1 · Answer 2 · answered Feb 22 '15 at 14:50

Obtaining that value requires analytic continuation when using the Zeta function.

It is necessary to understand that the sum of infinite natural numbers is different than finite sums. The moment you stop adding successive natural numbers, you get a huge positive value. It is also necessary to understand that by simply manipulating the series 1+2+3+4...and 1-2+3-4... , you can easily get -1/12. (Note: Be careful when treating infinite sums as finite)

Let a= 1+2+3+4+5+06+7+08+9+10.... Let 4a= 4 +8 +12 +16 .... Subtracting the second from the first, you get -3c = 1-2+3-4+5... (Subtract the 4 from the 2, the 8 from the four, and so on)

Now, -3c = 1-2+3-4+5... = 1/(1+x)^2 with x=1 (series formula) = 1/(1+1)^2 = 1/4

Therefore c=-1/12

This is a non-rigorous (slightly intuitive) way of understanding.

For the proof requiring zeta function, involving differentiation, see https://www.youtube.com/watch?v=w-I6XTVZXww

It's an excellent video, which takes care of the proof you wish to see, and I hope it helps you!

I'm pretty sure this answers your question regarding the proof. If you need more details don't hesitate to ask. — Panglossian Oporopolist, Feb 22 '15 at 14:51

Evaluating, $\sum_{n=1}^{\infty} n$

2 Answers2