# Finding a closed-form for the sum $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}$

Let $$\mathcal{S}$$ denote the sum of the following alternating series:

$$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}\approx-1.392562725547,$$

where $$H_{n}$$ denotes the $$n$$-th harmonic number.

Question: Is it possible to obtain a closed-form expression for $$\mathcal{S}$$ in terms of elementary functions and well-known special functions such as polylogarithms?

My approach was to use the following integral representation for the $$n$$-th harmonic number to convert the sum $$\mathcal{S}$$ into an integral via the technique of summing under the integral sign:

$$H_{n}=\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{n}}{1-t};~~~\small{n\in\mathbb{Z}_{\ge0}}.$$

We find

\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}\\ &=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}}\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{2n}}{1-t}\\ &=\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}}\cdot\frac{1-t^{2n}}{1-t}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{1-t}\left[\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}}-\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}t^{2n}}{n^{4}}\right]\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{4}{\left(-1\right)}-\operatorname{Li}_{4}{\left(-t^{2}\right)}}{1-t}\\ &=-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\operatorname{Li}_{3}{\left(-t^{2}\right)}}{t};~~~\small{I.B.P.}\\ &=2\operatorname{Li}_{2}{\left(1\right)}\operatorname{Li}_{3}{\left(-1\right)}-\int_{0}^{1}\mathrm{d}t\,\frac{4\operatorname{Li}_{2}{\left(t\right)}\operatorname{Li}_{2}{\left(-t^{2}\right)}}{t};~~~\small{I.B.P.}\\ &=-\frac32\zeta{\left(2\right)}\,\zeta{\left(3\right)}-4\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}\operatorname{Li}_{2}{\left(-t^{2}\right)}}{t}.\\ \end{align}

Any ideas how to proceed from here?

• Your question is a duplicate The calculation of the series $\sum_{n=1}^{\infty} (-1)^{(n-1)}\frac{H_{2n}}{n^4}$, I suggest one makes use of approach0 before asking. Jul 5, 2021 at 6:14
• @JorgeLayja the solutions in that post are cumbersome and unnecessarily involved. The one below is not only more efficient and self contained it generalises the calculation to arbitrary powers. So i'm not sure why i was downvoted. I'd suggest the moderators remove the old post and keep this one instead. Jul 5, 2021 at 7:25
• @mathstackuser12 I've read your proof, it is true, very nice approach! I'll upvote it, by the way I don't think the approaches found there are unnecessarily involved or cumbersome, Ali's approach was really elegant and only used elementary elements and techniques but it is true, your approach is indeed more self contained and efficient in a way. Jul 5, 2021 at 9:33

Consider $$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}}, n\ge 0$$ Then $$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k+1}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\pi f\left( k+\tfrac{1}{2} \right)}{2\left( z-\tfrac{2k+1}{2} \right)}+O\left( 1 \right)$$ On the other hand we also have $$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}f\left( k \right)}{2{{\left( z-k \right)}^{2}}}+\frac{{{\left( -1 \right)}^{k}}\left\{ {{H}_{2k}}f\left( k \right)+\tfrac{1}{2}f'\left( k \right) \right\}}{z-k}+O\left( 1 \right), k\ge 0$$ Similarly $$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{z\to -k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma \right)f\left( -k \right)}{z+k}, k>0$$ The only other residues are those due to $$f$$ which we assume has one pole at the origin of order at least $$2$$. The sum of residues over the entire plane is zero. Hence \begin{align} & \sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}{{H}_{2k}}f\left( k \right)} \\ & =-\frac{1}{2}\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}f'\left( k \right)}-\frac{\pi }{2}\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}f\left( k+\tfrac{1}{2} \right)}-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma \right)f\left( -k \right)}\\&-\underset{z=0}{\mathop{res}}\,\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)} \\ \end{align} Letting $$f\left( z \right)=\frac{1}{{{z}^{4}}}$$ we find \begin{align} & \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}} \\ & =-\frac{\pi }{192}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right) \right\}+\frac{2}{3}{{\pi }^{2}}\zeta \left( 3 \right)+\frac{113}{8}\zeta \left( 5 \right)-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\frac{{{H}_{2k-1}}}{{{k}^{4}}}} \\ \end{align} Now $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=A-\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}\Rightarrow \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( {{H}_{2k}}+{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$ But we can write this as $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( \frac{1}{2k}+2{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$ hence $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}=\frac{1}{2}A+\frac{15}{64}\zeta \left( 5 \right)$$ We have then $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=\frac{\pi }{384}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right) \right\}+\frac{{{\pi }^{2}}}{3}\zeta \left( 3 \right)+\frac{437}{64}\zeta \left( 5 \right)$$