How do you prove that $$\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\zeta(n) = \gamma$$ where $\gamma$ is the Euler-Macheroni constant? This series kind of appeared in one of the questions I asked earlier; you just need to do some rearranging to get to this series. Here is WolframAlpha calculating the series. I believe I have almost proved it (my calculations below), but I'm unsure at the end. I would also like to see if there is some other way of proving them (I don't think there is, but it would be cool).

My "proof" (not sure if it right):

\begin{align} \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\zeta(n) & =\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n}\sum_{k=1}^{\infty}\frac{1}{k^n} \\\\ & = \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{n}}{n}\frac{1}{k^n}\end{align}

Here I interchange the summations (is it possible to do so?):

\begin{align} \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{n}}{n}\frac{1}{k^n} & = \sum_{k=1}^{\infty}\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n}\frac{1}{k^n} \\\\ & = \sum_{k=1}^{\infty} \left(\frac{1}{k} + \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\frac{1}{k^n}\right)\end{align}

Recall the Taylor series for $\ln(x)$:

$$\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n$$

By plugging in $\frac{1}{x}$, changing $x$ to $k$ and multiplying by $-1$ on both sides we get: $$-\ln\left(\frac{k+1}{k}\right) = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\frac{1}{k^n}$$

...which is exactly what we need. So plugging the result into the series above we get:

\begin{align} \sum_{k=1}^{\infty} \left(\frac{1}{k} + \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\frac{1}{k^n}\right) & = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \ln\left(\frac{k+1}{k}\right)\right) \\\\ & = \sum_{k=1}^{\infty} \frac{1}{k} - \sum_{k=1}^{\infty}\ln\left(\frac{k+1}{k}\right) \end{align}

Recall the definition of the Euler-Macheroni constant: $$\gamma = \lim_{n\to\infty}(H_n - \ln(n))$$

Now clearly the term $\sum_{k=1}^{\infty} \frac{1}{k}$ is the $H_n$ part of the definition, but here's is where I get a little stuck; how does $$\sum_{k=1}^{\infty}\ln\left(\frac{k+1}{k}\right) = \lim_{k\to\infty}\ln(k)$$

Otherwise I think my proof is quite correct, but can anybody help at the end of it?

  • $\begingroup$ This is a telescopic series. \begin{align}\sum_{k=1}^n\ln{\left(\frac{k+1}{k}\right)}&=\sum_{k=1}^n(\ln{(k+1)}-\ln{(k)})\\&=\ln{(n+1)}\\\end{align} $\endgroup$ Commented May 1, 2020 at 10:21
  • $\begingroup$ @PeterForeman Alright thanks, I didn't realise I could apply it here since I haven't learnt it well enough yet. $\endgroup$ Commented May 1, 2020 at 10:25
  • $\begingroup$ @CasimirRönnlöf You're also fine to interchange the summations, though to be more precise you should separate out the summand when $ k = 1 $ so that you can actually use the Fubini theorem. As it stands, the summand is not $ L^1 $, because the harmonic series over $ n $ for $ k = 1 $ diverges. $\endgroup$
    – Ege Erdil
    Commented May 1, 2020 at 10:37
  • 3
    $\begingroup$ It can be seen as a consequence of this formula : $\ln \Gamma(1 + x) = - \gamma x + \sum_{n = 2}^{\infty} \frac {( - 1)^{n}}{n}\zeta(n) \, x^{n}$ with $x \to 1$ ; see here. $\endgroup$
    – Jean Marie
    Commented May 1, 2020 at 10:43
  • 1
    $\begingroup$ @JeanMarie Oh wow, that's cool, since I'm also quite interested in the Gamma function $\endgroup$ Commented May 1, 2020 at 10:47

1 Answer 1


We can't write

$$\sum_{k=1}^{\infty} \left(\frac{1}{k} - \ln\left(\frac{k+1}{k}\right)\right)=\sum_{k=1}^\infty\frac1k-\sum_{k=1}^\infty\ln\left(\frac{k+1}{k}\right)$$

because both of these two series are divergent. To fix this issue, we use the limit

$$\sum_{k=1}^{\infty} \left(\frac{1}{k} - \ln\left(\frac{k+1}{k}\right)\right) = \lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{1}{k} - \ln\left(\frac{k+1}{k}\right)\right)$$

$$=\lim_{n\to \infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=1}^n\ln\left(\frac{k+1}{k}\right)\right)$$

$$=\lim_{n\to \infty}\left(H_n-\ln(n+1)\right)$$

$$\overset{n+1=m}{=}=\lim_{m\to \infty}\left(H_{m-1}-\ln(m)\right)$$

$$=\lim_{m\to \infty}\left(H_m-\frac1m-\ln(m)\right)$$

$$=\lim_{m\to \infty}\left(H_m-\ln(m)\right)-\lim_{m\to \infty}\frac1m$$



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