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Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$

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Is it possible to generalize it to n^a instead of n^4?… – Piotr Shatalin May 8 '13 at 2:35
up vote 21 down vote accepted

$$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}=\frac{28}{243}+\frac{10}{81} \log \left(\frac{2}{3}\right).$$

Hint: Change the order of summation: $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}=\sum_{n=1}^\infty\sum_{k=1}^n\frac{(-1)^n n^4}{2^n k}=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{(-1)^n n^4}{2^n k}.$$

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How is this a hint? – Did Aug 26 '13 at 21:54

A related problem. Here is another approach. Recalling the generating function of the harmonic numbers

$$ \sum_{n=1}^{\infty} H_n x^n = \frac{\ln(1-x)}{x-1} \implies (xD)^4\sum_{n=1}^{\infty} H_n x^n=(xD)^4 \frac{\ln(1-x)}{x-1}, $$

$$ \implies \sum_{n=1}^{\infty} H_n n^4 x^n = {\frac {x \left( 1+11\,x+11\,{x}^{2}+{x}^{3} \right) \ln \left( 1-x \right) }{ \left( -1+x \right) ^{5}}}+{\frac {x \left( -1-27\,{x}^{2} -18\,x-4\,{x}^{3} \right) }{ \left( -1+x \right) ^{5}}}.$$

Substituting $x=-\frac{1}{2}$ in the above identity gives the desired result

$$ \sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}={\frac {28}{243}}-{\frac {10}{81}}\,\ln \left( \frac{3}{2} \right). $$

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Pretty nifty (+1). – Ron Gordon Aug 26 '13 at 20:39
@RonGordon: Thanks for the comment. – Mhenni Benghorbal Sep 3 '13 at 17:31
+1. That was the 'natural' way. – Felix Marin Sep 12 '14 at 18:13
@FelixMarin: Thanks for the comment. I really appreciate it. – Mhenni Benghorbal Sep 12 '14 at 18:24

Hint: Set $f_k=H_k$ and find a function $g_k$ such that $g_{k+1}-g_k=\frac{(-1)^kk^4}{2^k}$ and use summation by parts to get: $$\sum_{k=1}^nf_k(g_{k+1}-g_k)=f_{n+1}g_{n+1}-f_1g_1-\sum_{k=1}^n(f_{k+1}-f_k)g_{k+1}$$ $$\sum_{k=1}^nH_k(\frac{(-1)^kk^4}{2^k})=f_{n+1}g_{n+1}-f_1g_1-\sum_{k=1}^n(\frac{1}{k+1})g_{k+1}$$ Find the limit as $n\rightarrow \infty$. $g_k$ will be a function of the form $(\frac{-1}{2})^kp(k)+C$ for some polynomial $p$ and constant $C$. Thus, I believe the RHS will be a sum of the form: $$\sum_{k=1}^{\infty}[(\frac{-1}{2})^kp(k)+C]\frac{1}{k+1}$$ which will be just tedious to evaluate.

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