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The alternating sum $\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{\sqrt{n}}}=(1-\sqrt{2})\zeta(\frac{1}{2})$ according to Wolfram Alpha. I tried to derive this result myself, and found it easy using the naïve way of expanding $\zeta(\frac{1}{2})$ into its infinite sum $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots$ and then multiplying by $\frac{1}{\sqrt{2}}$ to get every even number, then simply subtracting 2 times this from $\zeta(\frac{1}{2})$ to get the final result. However, $\zeta(\frac{1}{2})$ diverges using the sum definition of zeta. I know there are extensions that allow you to calculate zeta for 1/2, however, I don't understand why using those extensions suddenly allows a diverging series to be used in intermediate expressions as if it converged. It feels like I'm doing $\infty-\infty$ and getting a reasonable result.

RobPratt
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girobuz
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    The zeta function isn't simply $\sum_{n\geq 1}\frac{1}{n^s}$ for all complex numbers, for that we use the analytic continuation, for example, it's clear that $\sum_{n\geq 1}\frac{1}{\sqrt{n}}$ diverges due to the integral test, but we can still define $\zeta(\frac{1}{2})$ – Tompad May 02 '25 at 19:14
  • Indeed, once you get the product formula for $\zeta,$ you will see the $\zeta(x)$ for real $x$ in $(0,1)$ can be defined roughly to make this true. – Thomas Andrews May 02 '25 at 19:31
  • This is https://oeis.org/A113024, obtained by analytic continuation of the definition of the eta-function. – R. J. Mathar May 02 '25 at 21:55

1 Answers1

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There is some analysis needed to prove this, but it can all be swept under the rug by leveraging known facts about the Dirichlet eta function $$ \eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}. $$ It is known (and pretty simple to prove when Dirichlet series are first studied) that:

  • the series defining $\eta(s)$ converges for all complex $s$ with $\Re s>0$;
  • if $\Re s>1$ then $\eta(s) = (1-2^{1-s})\zeta(s)$ (basically by the infinite-series procedure described in the OP);
  • by analytic continuation, $\eta(s) = (1-2^{1-s})\zeta(s)$ for all $s\ne1$ with $\Re s>0$.

With these background facts established, all we have to do is plug $s=\frac12$ into the last identity. Note that doing so isn't related to the infinite series $\sum_{n=1}^\infty 1/\sqrt n$ at all.

Greg Martin
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