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How can I prove that $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$

I think this post can help me, but I'm not sure.

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  • $\begingroup$ You like Mathematica, don't you? $\endgroup$ Commented Dec 16, 2013 at 13:47

4 Answers 4

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UPDATED: In this previous answer I proved that (if $\,\operatorname{Li}$ is the polylogarithm) : $$S(x):=\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=\zeta(3)+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)$$

Taking the limit as $x\to 1$ and $x\to -1$ we get : $$S(1)=\sum_{n=1}^\infty \frac{H_n}{n^2}=2\,\zeta(3),\quad S(-1)=\sum_{n=1}^\infty \frac{(-1)^n\;H_n}{n^2}=-\frac 58\zeta(3)$$

To obtain the limit for $S(1)$ use $\,\operatorname{Li}_n(\epsilon)\sim\epsilon\,$ as $|\epsilon|\to 0\,$ and $\,\operatorname{Li}_n(1)=\zeta(n)\,$.

For $S(-1)$ the method is rather long $(*)$ so let's come back to the integral expression from my previous answer applied to $x=-1$ (using $\displaystyle\operatorname{Li}_3(-1)=-\frac34\zeta(3)$ from $(*)$) : \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^2}(-1)^n&=\int_0^{-1} \frac {\ln(1-x)^2}{2\,x}dx+\operatorname{Li}_3(-1)\\ &=\int_0^1 \frac {\ln(1+x)^2}{2\,x}dx-\frac 34\zeta(3)\\ \end{align}

Integrals like $\;\displaystyle a_k:=\int_0^1 \frac {\ln(1+x)^k}xdx\,$ were studied by Nielsen and Ramanujan who found that $\displaystyle a_1=\frac{\zeta(2)}2,\;a_2=\frac{\zeta(3)}4$ so that $\;\displaystyle S(-1)=\frac 12\frac{\zeta(3)}4-\frac{\zeta(3)}4=-\frac 58\zeta(3)$

We conclude that : $$S(1)+S(-1)=2\sum_{m=1}^\infty \frac{H_{2m}}{(2m)^2}=2\,\zeta(3)-\frac 58\zeta(3)=\frac {11}8\zeta(3)$$ and \begin{align} \sum_{n=1}^\infty \frac{H_{2n+1}}{n^2}&=\sum_{n=1}^\infty \frac{H_{2n}+\frac 1{2n+1}}{n^2}\\ &=\frac {11}4\zeta(3)+\sum_{n=1}^\infty \frac 1{n^2(2n+1)}\\ &=\frac {11}4\zeta(3)+\sum_{n=1}^\infty \frac 1{n^2}-\frac 4{2n}+\frac 4{2n+1}\\ &=\frac {11}4\zeta(3)+\sum_{n=1}^\infty \frac 1{n^2}+4\sum_{n=2}^\infty\frac {(-1)^{n-1}}{n}\\ &=\frac {11}4\zeta(3)+\zeta(2)+4(\ln(1+1)-1)\\ \end{align}

Concerning the series $\displaystyle \sum_{n=1}^\infty\frac{H_{2n+k}}{n^2}\;$ we simply add individual terms (as $+\frac 1{2n+1}$ previousy) and the same method still works : expand the remaining terms in partial fractions and evaluate the series.

For $k=5$ (using computer algebra of course) I got the previously evaluated $\; \displaystyle\frac {11}4\zeta(3)+$

$$\sum_{n=1}^\infty\frac 1{n^2}\left(\frac 1{2n+1}+\frac 1{2n+2}+\frac 1{2n+3}+\frac 1{2n+4}+\frac 1{2n+5}\right)=\\\sum_{n=1}^\infty\frac{137}{60n^2}-\frac{5269}{1800\,n} + \frac 1{2(n + 1)} + \frac 1{8(n + 2)} + \frac 4{2n + 1} + \frac 4{9( 2n + 3)} + \frac 4{25(2n + 5)}=\\ \frac{1036}{225}\ln(2) + \frac{137}{60}\zeta(2) - \frac{298373}{54000} $$

For $k=4$ the additional terms are $\displaystyle +\frac{40}9\ln(2) + \frac{25}{12}\zeta(2) - \frac{2281}{432}$ that you may compare with your result.

All results obtained for $k>0$ could be written as $\; \displaystyle R(k)=\frac{11}4\zeta(3)+p_k\,\zeta(2)+q_k\,\ln(2)+r_k\;$ for every $k>0$ with $p_k,q_k,r_k\in\mathbb{Q}\,$.

$(*)$ Alternative method to evaluate $S(-1)$ (complicated, left for the record only) :

  • $\displaystyle\operatorname{Li}_3(-1)=-\frac34\zeta(3)$ from the second formula $\;\displaystyle\operatorname{Li}_n(-1)=-\left(1-\frac 1{2^{n-1}}\right)\zeta(n)$
  • the inversion formula $\;\displaystyle\operatorname{Li}_3(z)=\operatorname{Li}_3(1/z)+2\zeta(2)\ln(z)-\frac{i\pi}2\ln(z)^2-\frac 16\ln(z)^3\;$ for $z>1$ applied to $z=2$.
  • the formula $\;\displaystyle\operatorname{Li}_3(1/2)=\frac 78\zeta(3)-\frac{\zeta(2)\ln(2)}2+\frac 16\ln(2)^3\,$ to conclude that $\;\displaystyle\operatorname{Li}_3(2)=\frac 78\zeta(3)-\frac{i\pi}2\ln(2)^2+\frac{\pi^2}4\ln(2)$.
  • the same way obtain $\;\displaystyle\operatorname{Li}_2(2)=-\frac{\pi^2}4-\frac12(\ln(2)+i\pi)^2+\frac 12\ln(2)^2\;$ from $\;\displaystyle\operatorname{Li}_2(z)=-\operatorname{Li}_2(1/z)+2\zeta(2)-i\pi\ln(z)-\frac 12\ln(z)^2\;$ and the formula for $\operatorname{Li}_2(1/2)$.
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    $\begingroup$ Thanks very much. Why $S(1)-S(-1)=2\sum_{m=1}^\infty \frac{H_{2m}}{(2m)^2}$? $\endgroup$ Commented Dec 16, 2013 at 16:10
  • $\begingroup$ If we let $n=2m$ then $S(1)-S(-1)=0$. $\endgroup$ Commented Dec 16, 2013 at 16:15
  • $\begingroup$ You are right @ALGEAN I had two signs wrong since the odd terms had to cancel (corrected). $\endgroup$ Commented Dec 16, 2013 at 16:44
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    $\begingroup$ +1. For a different evaluation of $S(-1)$ that relies solely on manipulation of the summation, see robjohn's answer here. $\endgroup$ Commented Dec 16, 2013 at 23:56
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    $\begingroup$ @MikeSpivey: really neat trick from robjohn indeed (and an impressive thread!). Many thanks to share Mike, $\endgroup$ Commented Dec 17, 2013 at 0:08
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{H_{2n + 1} \over n^{2}}} & = 4\sum_{n = 1}^{\infty}{H_{2n + 1} \over \pars{2n}^{2}} = 4\sum_{n = 1}^{\infty}{H_{n + 1} \over n^{2}}\, {1 + \pars{-1}^{n} \over 2} = 2\sum_{n = 2}^{\infty}{H_{n} \over \pars{n - 1}^{2}}\, \bracks{1 - \pars{-1}^{n}}\tag{1} \end{align}

\begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{H_{2n + 1} \over n^{2}}} & = 2\sum_{n = 2}^{\infty}H_{n}\bracks{1 - \pars{-1}^{n}}\ \overbrace{\pars{-\int_{0}^{1}y^{n - 2}\,\,\ln\pars{y}\,\dd y}} ^{\ds{1 \over \pars{n - 1}^{2}}} \\[5mm] & = -2\int_{0}^{1}\ln\pars{y}\sum_{n = 2}^{\infty} \bracks{H_{n}\,y^{n} - H_{n}\,\pars{-y}^{n}}\,{\dd y \over y^{2}} \\[5mm] & = -2\int_{0}^{1}\ln\pars{y}\bracks{-2y - {\ln\pars{1 - y} \over 1 - y} + {\ln\pars{1 + y} \over 1 + y}}\,\,{\dd y \over y^{2}} \\[5mm] & = 2\int_{0}^{1}\ln\pars{y}\bracks{{\ln\pars{1 - y} \over \pars{1 - y}y^{2}} - {\ln\pars{1 + y} \over \pars{1 + y}y^{2}} + {2 \over y}}\,\dd y\tag{2} \end{align}
However, \begin{align} {\ln\pars{1 - y} \over \pars{1 - y}y^{2}} & = {\ln\pars{1 - y} \over y} + {\ln\pars{1 - y} \over y^{2}} + {\ln\pars{1 - y} \over 1 - y} \\[5mm] {\ln\pars{1 + y} \over \pars{1 + y}y^{2}} & = -\,{\ln\pars{1 + y} \over y} + {\ln\pars{1 + y} \over y^{2}} + {\ln\pars{1 + y} \over 1 + y} \end{align}
With these expressions, $\ds{\pars{2}}$ is reduced to: \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{H_{2n + 1} \over n^{2}}}\ &\ =\ \overbrace{2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y^{2}} \over y}\,\dd y} ^{\ds{\,\mathcal{I}_{1}}}\ +\ \overbrace{% 4\int_{0}^{1}\ln\pars{y}\bracks{% {1 \over y} - {\,\mathrm{arctanh}\pars{y} \over y^{2}}}\,\dd y} ^{\ds{\,\mathcal{I}_{2}}} \\[3mm] &\ +\ \underbrace{% 2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y} \over 1 - y}\,\dd y} _{\ds{\,\mathcal{I}_{3}}}\ -\ \underbrace{ 2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y} \over 1 + y}\,\dd y} _{\ds{\,\mathcal{I}_{4}}} \\[5mm] & = \,\mathcal{I}_{1} + \,\mathcal{I}_{2} + \,\mathcal{I}_{3} -\,\mathcal{I}_{4} \tag{3} \end{align}
We'll evaluate the integrals $\ds{\,\mathcal{I}_{k}\,,\ k = 1,2,3,4}$:

  1. $\ds{\large\,\mathcal{I}_{1} :\ ?}$. \begin{align} \,\mathcal{I}_{1} & = 2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y^{2}} \over y}\,\dd y\ \stackrel{y^{2}\ \mapsto\ y}{=}\ \half\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y} \over y}\,\dd y\tag{4.a} \\[5mm] & = -\,\half\int_{0}^{1}\ln\pars{y}\Li{2}'\pars{y}\,\dd y = \half\int_{0}^{1}{\Li{2}\pars{y} \over y}\,\dd y = \half\int_{0}^{1}\Li{3}'\pars{y}\,\dd y \\[5mm] & = \half\Li{3}\pars{1} = \fbox{$\ds{\ \half\,\zeta\pars{3}\ }$} = \,\mathcal{I}_{1}\tag{4.b} \end{align}
  2. $\ds{\large\,\mathcal{I}_{2} :\ ?}$. With the $\ds{\,\mathrm{arctanh}\pars{y}}$ expansion in powers of $\ds{y}$: \begin{align} \,\mathcal{I}_{2} & = 4\int_{0}^{1}\ln\pars{y}\bracks{% {1 \over y} - {\,\mathrm{arctanh}\pars{y} \over y^{2}}}\,\dd y = 4\int_{0}^{1}\ln\pars{y}\bracks{% {1 \over y} - {1 \over y^{2}}\sum_{n = 0}^{\infty}{y^{2n + 1} \over 2n + 1}}\,\dd y \\[5mm] & = -4\sum_{n = 0}^{\infty}{1 \over 2n + 3}\ \overbrace{% \int_{0}^{1}\ln\pars{y}\,y^{2n + 1}\,\,\dd y} ^{\ds{-\,{1 \over 4\pars{n + 1}^{2}}}}\ =\ \left.-\,\half\,\partiald{}{\mu} \sum_{n = 0}^{\infty}{1 \over \pars{n + \mu}\pars{n + 3/2}} \right\vert_{\ \mu\ =\ 1} \\[5mm] & = \left.\vphantom{\huge A^{A}}-\,\half\,\partiald{}{\mu} \bracks{\Psi\pars{\mu} - \Psi\pars{3/2} \over \mu - 3/2} \right\vert_{\ \mu\ =\ 1}\qquad\qquad\pars{~\Psi:\ Digamma\ Function~} \\[5mm] & = \fbox{$\ds{\ 4\ln\pars{2} - 4 + {\pi^{2} \over 6}\ }$} = \,\mathcal{I}_{2}\tag{5} \end{align}
  3. $\ds{\large\,\mathcal{I}_{3} :\ ?}$. \begin{align} \,\mathcal{I}_{3} & = 2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y} \over 1 - y}\,\dd y\ \stackrel{y\ \mapsto\ \pars{1 - y}}{=}\,\,\, 2\int_{0}^{1}{\ln\pars{1 - y}\ln\pars{y} \over y}\,\dd y \\[5mm] & = 4\,\mathcal{I}_{1}\qquad\qquad \pars{~\mbox{see}\ \pars{\mbox{4.a}}\ \mbox{and}\ \pars{\mbox{4.b}}~} \\[5mm] & = \fbox{$\ds{\ 2\zeta\pars{3}\ }$} = \,\mathcal{I}_{3}\tag{6} \end{align}
  4. $\ds{\large\,\mathcal{I}_{4} :\ ?}$. \begin{align} \,\mathcal{I}_{4} & = 2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y} \over 1 + y}\,\dd y = -\int_{0}^{1}{\ln^{2}\pars{1 + y} \over y}\,\dd y \\[5mm] & \stackrel{\pars{1 + y}\ \mapsto\ y}{=}\ -\int_{1}^{2}{\ln^{2}\pars{y} \over y - 1}\,\dd y\ \stackrel{y\ \mapsto\ 1/y}{=}\ -\int_{1}^{1/2}{\ln^{2}\pars{1/y} \over 1/y - 1}\,\pars{\dd y \over -y^{2}} \\[5mm] & = -\int_{1/2}^{1}{\ln^{2}\pars{y} \over y\pars{1 - y}}\,\dd y = -\int_{1/2}^{1}{\ln^{2}\pars{y} \over y}\,\dd y - \int_{1/2}^{1}{\ln^{2}\pars{y} \over 1 - y}\,\dd y \\[5mm] & = -\,{1 \over 3}\,\ln^{3}\pars{2} + \ln^{3}\pars{2} - \int_{1/2}^{1}\ln\pars{1 - y}\,2\ln\pars{y}\,{1 \over y}\,\dd y \\[5mm] & = {2 \over 3}\,\ln^{3}\pars{2} + 2\int_{1/2}^{1}\Li{2}'\pars{y}\ln\pars{y}\,\dd y \\[5mm] & = {2 \over 3}\,\ln^{3}\pars{2} + 2\ln\pars{2}\Li{2}\pars{\half} -2\int_{1/2}^{1}\overbrace{{\Li{2}\pars{y}\over y}}^{\ds{\Li{3}'\pars{y}}} \,\dd y \\[5mm] & = {2 \over 3}\,\ln^{3}\pars{2} + 2\ln\pars{2}\Li{2}\pars{\half} -2\Li{3}\pars{1} + 2\Li{3}\pars{\half} \end{align} The $\ds{\Li{s}}$ values involved in this expression are well known: $\ds{\Li{2}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}$, $\ds{\Li{3}\pars{1} = \zeta\pars{3}}$ and $\ds{\Li{3}\pars{\half} = {7 \over 8}\,\zeta\pars{3} + {1 \over 6}\,\ln^{3}\pars{2} - {\pi^{2} \over 12}\,\ln\pars{2}}$ such that \begin{equation} \,\mathcal{I}_{4} = \fbox{$\ds{\ -\,{1 \over 4}\,\zeta\pars{3}\ }$}\tag{7} \end{equation}
    With $\ds{\pars{3}}$$,\ds{\pars{\mbox{4.b}}}$, $\ds{\pars{5}}$, $\ds{\pars{6}}$ and $\ds{\pars{7}}$, we find the $\ds{\underline{final\ result}}$ $$ \color{#f00}{\sum_{n = 1}^{\infty}{H_{2n + 1} \over n^{2}}} = \color{#f00}{{11 \over 4}\,\zeta\pars{3} + \zeta\pars{2} + 4\ln\pars{2} - 4} \approx 3.7232\,,\qquad {\pi^{2} \over 6} = \zeta\pars{2} $$
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Using the following nice rule: $$\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}\left(\frac{1+(-1)^n}{2}\right)$$

We get \begin{align} S&=\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n+1}}{{(2n)}^2}=4\left(\frac12\sum_{n=1}^\infty\frac{H_{n+1}}{n^2}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n+1}}{n^2}\right)\\ &=2\sum_{n=1}^\infty\frac{H_n}{n^2}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}+2\sum_{n=1}^\infty\frac{1}{n^2(n+1)}+2\sum_{n=1}^\infty\frac{(-1)^n}{n^2(n+1)}\\ &=2\left(2\zeta(3)\right)+2\left(-\frac58\zeta(3)\right)+2\left(\zeta(2)-1\right)+2\left(2\ln2-\frac12\zeta(2)-1\right)\\ &=\frac{11}4\zeta(3)+\zeta(2)+4\ln2-4 \end{align}

Note that $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}$ was obtained from using the generating function where we set $x=-1$ :

$$\sum_{n=1}^\infty x^n\frac{H_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

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Different approach:

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

where $$\sum_{n=1}^\infty\frac{H_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}$$ $$=4\sum_{n=1}^\infty\frac{1}{2n}\left(-\int_0^1x^{2n-1}\ln(1-x)dx\right)$$

$$=-2\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{x^{2n}}{n}$$

$$=2\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}dx$$

$$=2\int_0^1\frac{\ln^2(1-x)}{x}dx+2\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$

$$=2(2\zeta(3))+2(-\frac58\zeta(3))=\frac{11}{4}\zeta(3)$$

and

$$\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{2n}dx$$ $$=\int_0^1\sum_{n=1}^\infty \frac{x^{2n}}{n^2}=\int_0^1\text{Li}_2(x^2)dx$$

$$=x\text{Li}_2(x^2)|_0^1+2\int_0^1\ln(1-x^2)dx$$

$$=\zeta(2)+2(x-1)\ln(1-x^2)|_0^1-4\int_0^1\frac{x}{1+x}dx$$

$$=\zeta(2)-4(1-\ln(2))$$

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