It's well known that $\zeta(0)=\sum{\frac{1}{n^0}}=\sum{1}=-\frac{1}{2}$, so I know there is something wrong with extending the method Mathologer described here infinetely many times:

Partial sums of $\sum{1}$:

$1, 2, 3, 4, 5, 6...$

Partial averages:

$\frac{2}{2}, \frac{3}{2}, \frac{4}{2}, \frac{5}{2}...$

Partial averages of that:

$\frac{4}{4}, \frac{5}{4}, \frac{6}{4}, \frac{7}{4}...$

It can be easily seen that the partial average after you applied this procedure $n$ times is $\frac{2^n}{2^n}, \frac{2^n+1}{2^n}, \frac{2^n+2}{2^n}, \frac{2^n+3}{2^n}...$

So at any point, the average of the partial sum is $\frac{2^n+c}{2^n}$ with $c \in \mathbb{N}$.

For any $c$ we get $\lim_{n \to \infty}\frac{2^n+c}{2^n}=1$

My question is, where exactly does the error occur? Can't we apply this procedure infinetely many times (but only arbitrarily many times)? If so, why not? Or is the error somewhere else?


closed as unclear what you're asking by Arnaud D., Rohan, Shailesh, C. Falcon, Leucippus Jan 9 '17 at 1:56

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  • $\begingroup$ There is actually something wrong in your procedure: you can apply the process of taking averages n times, and then take the limit, but you must apply the limit on n at the limit values of the averages, and they are all infinity: for example, if you apply the average one time, you get the sequence $1,\frac{3}{2}, 2,...$ whose limit is infinity; applying the process again, the limit of the sequence of the averages is again infinity; and so on, you always get the limit infinity. In other words, you first need to take the limit $c-> \infty$ and then $n -> \infty$, and not the reverse. $\endgroup$ – Klaramun Jan 8 '17 at 16:06
  • $\begingroup$ In order for it to be cesaro sumable, it needs to be have no constant part. So it need to look like $ e^{2 i nc \pi}* f(n)$ $\endgroup$ – Gerben Jan 17 '17 at 22:46

There is no error. It is simply the case that the following divergent series:


is not Cesàro summable.



which is a common misunderstanding. If one insists on approaching the problem this way, then you should use the Dirichlet eta function:


where $\eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s}$. Cesàro summing this for $s\le0$ will come out right. For a clear closed form, one should apply the Euler summation formula instead.

  • $\begingroup$ Nice exposition (+1)! Typo - it must read: $\zeta(0)=- \frac 12 \ne \cdots $ $\endgroup$ – Gottfried Helms Jan 12 '17 at 16:53
  • $\begingroup$ @GottfriedHelms ah yes, thanks for the catch. $\endgroup$ – Simply Beautiful Art Jan 12 '17 at 17:02

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