26
$\begingroup$

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: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$

$\endgroup$
23
$\begingroup$

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

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

$\endgroup$
  • 3
    $\begingroup$ How is this a hint? $\endgroup$ – Did Aug 26 '13 at 21:54
13
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ Pretty nifty (+1). $\endgroup$ – Ron Gordon Aug 26 '13 at 20:39
  • $\begingroup$ @RonGordon: Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Sep 3 '13 at 17:31
  • $\begingroup$ @FelixMarin: Thanks for the comment. I really appreciate it. $\endgroup$ – Mhenni Benghorbal Sep 12 '14 at 18:24
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.