Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$

Question

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.

Answer 1

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}

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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 $p_k,q_k,r_k\in\mathbb{Q}\,$.

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$(*)$ 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 $\operatorname{Li}_2(1/2)$.

Answer 2

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

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

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

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

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