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Since $\zeta(-1)=\frac{1}{1^{-1}}+\frac{1}{2^{-1}}+\frac{1}{3^{-1}}+\cdots=-\frac{1}{12}$, why do we still say that $\sum^\infty_{n=1}n\rightarrow+\infty$?

dcstup
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    Because $\zeta(-1)\ne\sum_{n=1}^\infty n$. – Eclipse Sun Sep 24 '15 at 06:19
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    $\zeta(-1)$ is not $\frac{1}{1^{-1}}+\frac{1}{2^{-1}}+\cdots$. The equation $\zeta(s) = 1^{-s}+2^{-s}+\cdots$ is only valid for $Re(s)>1$. – pregunton Sep 24 '15 at 06:20
  • You might find this blog post I wrote once about this question useful. – davidlowryduda Sep 24 '15 at 06:26
  • This question appears regularely and the "conclusion" seems to have an influence to the string theory. The tragedy : Mathematics is necessary to quantify physical results. But unfortunately, modern physics tries to interpret mathematical results (especially, if they are "beautiful") in a physical way. But Mathematics can, of course, not replace the understanding of physics. – Peter Sep 24 '15 at 12:31
  • This leads to tremendous problems, but instead of rethinking about this method, the problems turn into sensational physical "realities". This is especially the case in astronomics. – Peter Sep 24 '15 at 12:34
  • Note the linked question with 172 upvotes ! The discussion seems to result in the acceptance of the above equation because it can be derived from the extension of the riemann-zeto-function. I agree with Daniel Fischer, who calls this a "misused result". – Peter Sep 24 '15 at 12:38

1 Answers1

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First of all, it is NOT TRUE that

$$\zeta(-1) = \frac{1}{1^{-1}} + \frac{1}{2^{-1}} + \dots$$

The Riemann Zeta function is only defined as

$$\zeta(s) = \sum_{i=1}^\infty \frac{1}{n^s}$$

for values of $s$ for which the real part of $s$ is greater than one.

For other values, we can calculate the analytic continuation of $\zeta$ to define the function on the entire complex plane, but that means that we are expanding the definition beyond what it originally was.

You need to understand where our definitions come from:

  • First, we have infinite sequences.
  • From infinite sequences, we construct infinite series, and we define when these series converge.
  • Then, we take a very specific family of infinite series, and we construct a function that sums them.
  • Then, we expand the definition of this function to other numbers.

Now, if we take a series and ask "does it converge", we cannot say "yes because Riemann's zeta says so". We have an earlier definition, and to answer convergence, we need to use that definition.

So, why does $$\sum_{i=0}^\infty i$$

diverge? Simply. Take the original definition of convergence of series, and the answer becomes:

Because $$\sum_{i=0}^\infty a_i$$

is defined as $$\lim_{N\to\infty} \sum_{i=0}^N a_i$$

And the sequence in the case when $a_i = i$ diverges toward $\infty.$

5xum
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    +1 Upvoted to counter vandal (?) downvote. – Brian Tung Sep 24 '15 at 06:15
  • I am the downvoter. This does not answer the question at all. –  Sep 24 '15 at 06:16
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    @avid19 This perfectly answers the question of why the sum is $\infty$. I intentionally left out any mention of Ramanujan's summation and Riemann's zeta functions because they have nothing to do with the fact that the sum is $\infty.$ – 5xum Sep 24 '15 at 06:16
  • @5xum that isn't quite his/her question though. –  Sep 24 '15 at 06:17
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    An answer should use the phrase "analytic continuation". –  Sep 24 '15 at 06:18
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    @avid19 No. If the question was "Why is $\zeta(-1) = -\frac{1}{12}$" ten the answer should involve that phrase. Anyway, the question was changed, and I will now edit my answer. – 5xum Sep 24 '15 at 06:19
  • I upvoted your answer after you addressed the source of confusion. Before it didn't address OP's confusion at all. –  Sep 24 '15 at 06:27
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    @avid19 I know. That was my intent. I wanted OP to clearly show, in his question, what he did wrong. I did not want to assume where his confusion came from. I hoped OP would change the question. And I think the original answer was incomplete, yes, but not completely wrong, as you say. But anyway, we don't agree, and this is the internet, so nothing surprising. – 5xum Sep 24 '15 at 06:29
  • @5xum How can we possibly sleep tonight knowing some stranger on the internet might be wrong? :) –  Sep 24 '15 at 06:31
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    @avid19 Personally, I will sleep on mi right side covered on a blanket. How will you sleep? – 5xum Sep 24 '15 at 06:32
  • @5xum Again with missing the point. :) –  Sep 24 '15 at 06:36