Evaluating $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$

Question

I would appreciate to understand the main steps giving the evaluation of this series: $$ S=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$$ where $H_n$ is the harmonic number. I've tried with no success to obtain this sum with the help of Wolfram Alpha.

Answer 1

Using integral representation for harmonic number given by Leonhard Euler \begin{equation} H_k=\int_0^1\frac{1-x^k}{1-x}\,dx \end{equation} and Taylor series for natural logarithm \begin{equation} \ln(1+y)=\sum_{k=1}^\infty\, (-1)^{k+1}\frac{x^k}{k}\qquad,\qquad\mbox{for}\,\,|y|<1 \end{equation} we have \begin{align} S&=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}\\ &=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_0^1\frac{1-x^{2n}}{1-x}\,dx\\ &=\int_0^1\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left[1-x^{2n}\right]\,\frac{dx}{1-x}\\ &=\int_0^1\left[\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\right]\,dx\\ \end{align} Now, apply integration by parts by taking $u=\ln2-\ln\left(1+x^2\right)$ and $dv=\dfrac{dx}{1-x}$. We obtain \begin{equation} S=-\lim_{t\to1}\left.\left[\ln2-\ln\left(1+x^2\right)\right]\ln(1-x)\right|_0^t-2\int^1_0\frac{x\ln(1-x)}{1+x^2}\,dx \end{equation} Since we have an indeterminate form $0\cdot\infty$ as $t\to1$, we may write our limit as \begin{equation} -\lim_{t\to1}\frac{\ln(1-t)}{\frac{1}{\ln2-\ln\left(1+t^2\right)}} \end{equation} then apply L'Hospital's rule twice to obtain \begin{equation} \lim_{t\to1}\left[\ln2-\ln\left(1+t^2\right)\right]\ln(1-t)=0 \end{equation} Therefore, we may write our original sum as \begin{equation} S=-2\int^1_0\frac{x\ln(1-x)}{1+x^2}\,dx \end{equation} The evaluation of the latter integral can be seen here. I take no credit for it. There are some good answers, but my personal choice is the answer using differentiation under the integral sign method since it's very simple and easy to understand. Anyway, this is a good question, thanks for sharing. ツ

Answer 2

Here is an approach to obtain

$$ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}=\frac{5\pi^2}{48}-\frac14 \ln^22. \tag1 $$

Recall the standard series expansion $$ \log(1+z)= \sum_{n=1}^{\infty} (-1)^{n-1}\frac{z^{n}}{n},\quad |z|\leq 1, z\neq -1. \quad (*) $$ Then, by the Cauchy product, one has $$ \begin{align} \log(1+z)\log(1-z) &=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{z^{n}}{n}\sum_{n=1}^{\infty} \frac{-z^{n}}{n}\\ &=\sum_{n=1}^{\infty}\left( \sum_{k=1}^{n-1} \frac{(-1)^{k}}{k(n-k)}\right)z^{n}\\ &=\sum_{n=1}^{\infty}\!\left( \frac{1+(-1)^{n}}{n}\sum_{k=1}^{n-1} \frac{(-1)^{k}}{k}\right)\!z^{n}\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{k=1}^{2n-1} \frac{(-1)^{k}}{k}z^{2n}\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\!\left( H_{2n}-H_n-\frac1n\right)\!z^{2n}\\ \end{align} $$ equivalently $$ \begin{align} \sum_{n=1}^{\infty}\frac{ H_{2n}}{n}z^{2n}=\log(1+z)\log(1-z)+\sum_{n=1}^{\infty}\frac{ H_{n}}{n}z^{2n}+\sum_{n=1}^{\infty}\frac{ 1}{n^2}z^{2n} \end{align} \tag2 $$ Now, if you put $z:=i\,$ ($i^2=-1$) in $(2)$ you easily conclude with $$ \begin{align} & \log(1+i)=\log \left(\sqrt{2}\:e^{i\pi/4} \right)=\frac12 \ln2 + i\frac\pi 4 \\ & \log(1+i)\log(1-i)=\frac14 \ln^22+\frac{\pi^2}{16}\\ & \sum_{n=1}^{\infty}(-1)^{n}\frac{H_{n}}{n} =\frac{\ln^22}{2}-\frac{\pi^2}{12} \tag3\\ & \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^2} =-\frac{\pi^2}{12} \tag4 \end{align} $$ giving the desired result.

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We may prove $(3)$ by considering the standard identity, easily obtained from $(*)$, $$\sum_{n=1}^{\infty}(-1)^{n}H_n x^{n-1} = -\dfrac{\ln(1+x)}{x(1+x)} \quad -1 < x<1,\,x\neq0,$$ integrating from $x=0^+$ to $x=1^-$ getting $$\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} \!= \!- \int_{0}^{1}\dfrac{\ln(1+x)}{x(1+x)} dx = \!-\int_{0}^{1}\left(\dfrac{\ln(1+x)}{x}\! - \!\dfrac{\ln(1+x)}{1+x}\right) \! dx= -\dfrac{\pi^2}{12}+\dfrac{\ln^2 2}{2}$$ where we have classically used $(*)$ again to obtain $$\int_{0}^{1}\dfrac{\ln(1+x)}{x}dx = \dfrac{\pi^2}{12}. $$

Answer 3

From the generating function for the harmonic numbers, we immediately have that

$$\sum_{n=1}^{\infty} \frac{H_n}{n}z^n=\frac{1}{2}\log^2(1-z)+\operatorname{Li_2}(z)$$ and after multiplying both sides by $-2$ and setting $z=i$, we get $$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n}}{n}=-2\left(\frac{1}{2}\log^2(1-i)+\operatorname{Li_2}(i)\right)$$ $$=\frac{5\pi^2}{48}-\frac14 \ln^22$$

Q.E.D.

Answer 4

Here is another approach:

We can see that for every $n≥k$, $\frac{1}{2k-1}+\frac{1}{2k}$ is included in $H_{2n}$. So we may conclude that: $$ S=\sum_{k=1}^{\infty} \left[\left(\frac{1}{2k-1}+\frac{1}{2k}\right)\cdot\sum_{n=k}^{\infty} \frac{(-1)^{n-1}}{n}\right] $$ Now we have: $$ \sum_{n=k}^{\infty} \frac{(-1)^{n-1}}{n}=\sum_{n=k}^{\infty} \left[(-1)^{n-1}\int_0^1 x^{n-1}dx\right]=\int_0^1 \left[\sum_{n=k}^{\infty} (-x)^{n-1}\right]dx=\int_0^1 \frac{(-x)^{k-1}}{1+x}dx $$ And therefore: $$ S=\sum_{k=1}^{\infty} \left[\left(\frac{1}{2k-1}+\frac{1}{2k}\right)\cdot\int_0^1 \frac{(-x)^{k-1}}{1+x}dx\right]=\int_0^1 \frac{\sum_{k=1}^{\infty}\frac{(-x)^{k-1}}{2k-1}+\sum_{k=1}^{\infty}\frac{(-x)^{k-1}}{2k}}{1+x}dx $$ And now: $$ \arctan(x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^{2k-1}}{2k-1} \implies \sum_{k=1}^{\infty}\frac{(-x)^{k-1}}{2k-1}=\frac{\arctan\left(\sqrt x\right)}{\sqrt x} $$ $$ \ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k} \implies \sum_{k=1}^{\infty}\frac{(-x)^{k-1}}{2k}=\frac{\ln(1+x)}{2x} $$ Thus: $$ S=\int_0^1 \frac{\frac{\arctan\left(\sqrt x\right)}{\sqrt x}+\frac{\ln(1+x)}{2x}}{1+x}dx=\int_0^1 \frac{\arctan\left(\sqrt x\right)}{\sqrt x(1+x)}dx+\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x(1+x)}dx=\int_0^1 \frac{\arctan\left(\sqrt x\right)}{\sqrt x(1+x)}dx+\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}-\frac{\ln(1+x)}{1+x}dx=\int_0^1 \frac{\arctan\left(\sqrt x\right)}{\sqrt x(1+x)}dx+{\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}}-\frac{1}{2}\int_0^1\frac{\ln(1+x)}{1+x}dx $$ Now we can calculate the integrals. In the first one we can make the substitution $x\to \tan^2(x)$: $$ \int_0^1 \frac{\arctan\left(\sqrt x\right)}{\sqrt x(1+x)}dx=\int_0^{\frac{\pi}{4}} 2(\tan(x)(1+\tan^2(x)))\frac{x}{\tan(x)(1+\tan^2(x))}dx=2\int_0^{\frac{\pi}{4}} x\space dx=\frac{\pi^2}{16} $$ The second integral can be calculated as follows: $$ \int_0^1 \frac{\ln(1+x)}{x}dx=\int_0^1 \sum_{k=1}^\infty \frac{(-1)^{k-1}x^{k-1}}{k}dx=\sum_{k=1}^\infty \left[\frac{(-1)^{k-1}}{k}\int_0^1 x^{k-1} dx\right]=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2}=\sum_{k=1}^\infty \frac{1}{(2k-1)^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}=\sum_{k=1}^\infty \frac{1}{k^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}=\sum_{k=1}^\infty \frac{1}{k^2}-\frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^2}-\frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^2}=\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{12} $$ The third we may calculate with partial integration: $$ \int_0^1\frac{\ln(1+x)}{1+x}dx=\left[\ln^2(1+x)\right]_0^1-\int_0^1\frac{\ln(1+x)}{1+x}dx \iff \int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{1}{2}\left[\ln^2(1+x)\right]_0^1=\frac{\ln^2(2)}{2} $$ So finally we have: $$ S=\int_0^1 \frac{\arctan\left(\sqrt x\right)}{\sqrt x(1+x)}dx+{\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}}-\frac{1}{2}\int_0^1\frac{\ln(1+x)}{1+x}=\frac{\pi^2}{16}+\frac{1}{2}\frac{\pi^2}{12}-\frac{1}{2}\frac{\ln^2(2)}{2}=\frac{5\pi^2}{48}-\frac{\ln^2(2)}{4} $$

Answer 5

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{S \equiv \sum_{n\ =\ 1}^{\infty}\pars{-1}^{n - 1}\,{H_{2n} \over n}}$

With the Harmonic Number Generating Function $$ -\,{\ln\pars{1 - x} \over 1 - x} = \sum_{n\ =\ 1}^{\infty}H_{n}x^{n} $$

we'll have: \begin{align} -\,{\ln\pars{1 - x} \over 1 - x} &= \sum_{n\ =\ 1}^{\infty}H_{n}x^{n} \\ -\,{\ln\pars{1 + x} \over 1 + x} &= \sum_{n\ =\ 1}^{\infty}H_{n}\pars{-x}^{n} \\[5mm] -\,{\ln\pars{1 - x} \over 1 - x} -{\ln\pars{1 + x} \over 1 + x} &= 2\sum_{n\ =\ 1 \atop n\ \mbox{even}}^{\infty}H_{n}x^{n} =2\sum_{n\ =\ 1}^{\infty}H_{2n}x^{2n} \\[5mm] -\int_{0}^{x}\bracks{{\ln\pars{1 - t} \over 1 - t} +{\ln\pars{1 + t} \over 1 + t}}\,{\dd t \over t} &=\sum_{n\ =\ 1}^{\infty}H_{2n}\,{x^{2n} \over n} \end{align}

We'll set $\ds{x = \ic}$ such that \begin{align} S&=\color{#66f}{\large\sum_{n\ =\ 1}^{\infty}\pars{-1}^{n - 1}\,{H_{2n} \over n}} =\int_{0}^{\ic}\bracks{{\ln\pars{1 - t} \over 1 - t} +{\ln\pars{1 + t} \over 1 + t}}\,{\dd t \over t} \\[5mm]&=\sum_{\sigma\ =\ \pm}\int_{0}^{\ic} {\ln\pars{1 - \sigma t} \over \pars{1 - \sigma t}t}\,\dd t =\sum_{\sigma\ =\ \pm}\bracks{\sigma\int_{0}^{\ic} {\ln\pars{1 - \sigma t} \over 1 - \sigma t}\,\dd t +\int_{0}^{\ic}{\ln\pars{1 - \sigma t} \over t}\,\dd t} \\[5mm]&=\sum_{\sigma\ =\ \pm}\bracks{-\,\half\,\left.\ln^{2}\pars{1 - \sigma t} \right\vert_{t\ =\ 0}^{t\ =\ \ic} +\int_{0}^{\sigma\ic}{\ln\pars{1 - t} \over t}\,\dd t} \\[5mm]&=\sum_{\sigma\ =\ \pm}\bracks{-\,\half\,\ln^{2}\pars{1 - \sigma\ic} -\int_{0}^{\sigma\ic}{\rm Li}_{2}'\pars{t}\,\dd t} \\[5mm]&=-\,\Re\bracks{\ln^{2}\pars{1 - \ic} + 2\,{\rm Li}_{2}\pars{\ic}} =\color{#66f}{\large -\,\Re\ln^{2}\pars{1 - \ic} - 2\,\Re{\rm Li}_{2}\pars{\ic}} \qquad\pars{1} \end{align}

However, \begin{align} \color{#00f}{\Re\ln^{2}\pars{1 - \ic}} &=\Re\braces{\bracks{\half\,\ln\pars{2} - {\pi \over 4}\,\ic}^{2}} =\color{#00f}{{1 \over 4}\,\ln^{2}\pars{2} - {\pi^{2} \over 16}} \\[1cm] \color{#00f}{\Re{\rm Li}_{2}\pars{\ic}}& =\Re\sum_{n\ =\ 1}^{\infty}{\ic^{n} \over n^{2}} =\sum_{n\ =\ 1}^{\infty}{\ic^{2n} \over \pars{2n}^{2}} ={1 \over 4}\sum_{n\ =\ 1}^{\infty}{\pars{-1}^{n} \over n^{2}} \\[5mm]&={1 \over 4}\bracks{-\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2}} +\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}}} \\[5mm]&={1 \over 4}\bracks{-\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}} + \sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}} +\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}}} =-\,{1 \over 8}\ \underbrace{\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}} _{\ds{\color{#c00000}{\pi^{2} \over 6}}}\ =\ \color{#00f}{-\,{\pi^{2} \over 48}} \end{align}

With the previous partial result ( see expression $\pars{1}$ ) and these ones: \begin{align} S &\equiv \color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}\pars{-1}^{n - 1}\,{H_{2n} \over n}} =-\bracks{{1 \over 4}\,\ln^{2}\pars{2} - {\pi^{2} \over 16}} - 2\pars{-\,{\pi^{2} \over 48}} \\[5mm]&=\color{#66f}{\large{5\pi^{2} \over 48} - {1 \over 4}\,\ln^{2}\pars{2}} \approx {\tt 0.9080} \end{align}