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How to compute the following convergent series? or some hint! $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log n}{n}.$$ The Wolfram MATHEMATICA9.0 gives the result is $1/2(\log 2)^2-\gamma\log 2$, where $\gamma$ is tha Euler constant!

Matthew Conroy
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Riemann
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  • Do you want to show that the series is convergent? Or simply writing out the series expansion? – Rumplestillskin Jun 22 '17 at 03:14
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    @Riemann Are you sure that it's possible? – Michael Rozenberg Jun 22 '17 at 03:14
  • do you wish to check if it is divergent\convergent? hint: use Leibniz test for alternating series. – segevp Jun 22 '17 at 03:14
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    of course this series is convergent, what I want is its sum! – Riemann Jun 22 '17 at 03:16
  • check with matlab! I think you can't really know what the sum is. the more you compute the more accurate your sum is but I don't think it is possible otherwise. – segevp Jun 22 '17 at 03:22
  • Look here for some hints -> https://math.stackexchange.com/a/100894/254075. Start with the Dirichlet eta function $\eta(s):=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n^s}$ which is very close to the Riemann zeta function. Then use the Laurent series of the zeta function to help evaluate the derivative of the eta function at $s=1$ – sharding4 Jun 22 '17 at 03:38
  • @sharding4 I've posted a solution that evaluates the sum directly using the Euler-Maclaurin Summation Formula. – Mark Viola Jun 22 '17 at 04:09
  • Another chance is to represent $\log(n)$ as an integral through Frullani's theorem, exploit the Taylor series of $-\log(1-x)$ and recognize/prove a well-known integral representation for $\gamma$. – Jack D'Aurizio Jun 22 '17 at 04:54
  • @Riemann for a alternative series to converging a finite specific value we need the absolute convergence, here the absolute sum does not seem to be converging. (Rearrangements) – MAN-MADE Jun 22 '17 at 06:17
  • See also this question for the integral expression of the series – Yuriy S Jul 20 '19 at 06:53

2 Answers2

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We can write

$$\begin{align} \sum_{n=1}^{2N}(-1)^{n-1}\frac{\log(n)}{n}&=\sum_{n=1}^{2N} \frac{\log(n)}{n}-2\sum_{n=1}^N\frac{\log(2n)}{2n}\\\\ &=\sum_{n=N+1}^{2N}\frac{\log(n)}{n}-\log(2)\sum_{n=1}^N \frac1n \end{align}$$

Using the Euler-Maclaurin Summation Formula, we have

$$\begin{align} \sum_{n=}^{2N}(-1)^{n-1}\frac{\log(n)}{n}&=\int_{N+1}^{2N}\frac{\log(x)}{x}\,dx-\log(2)\,H_n+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12\log^2(2N)-\frac12\log^2(N+1)-\log(2)\log(N)-\log(2)\gamma+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12 \log^2(2)-\log(2)\gamma+O\left(\frac{\log(N)}{N}\right)\\\\ \end{align}$$

Taking the limit as $N\to \infty$ yields the coveted limit

$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log(n)}{n}=\frac12\log^2(2)-\log(2)\gamma$$

Mark Viola
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n}:\ {\large ?}}$.

Hereafter, I'll use well known identities which are related to the Riemann Zeta Function: \begin{align} &\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n} = \left.\partiald{}{\mu}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{1 - \mu}} \right\vert_{\ \mu\ =\ 0^{+}} = \left.\partiald{\bracks{\pars{1 - 2^{\mu}}\zeta\pars{1 - \mu}}}{\mu}\, \right\vert_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \partiald{}{\mu} \braces{\bracks{-\ln\pars{2}\mu - {1 \over 2}\,\ln^{2}\pars{2}\,\mu^{2} + \,\mrm{O}\pars{\mu^{3}}}\bracks{-\,{1 \over \mu} + \gamma + \,\mrm{O}\pars{\mu^{1}}}}_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \partiald{}{\mu}\braces{\ln\pars{2} + \bracks{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}}\mu + \,\mrm{O}\pars{\mu^{2}}}_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \bbx{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}} \end{align}

Felix Marin
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