We have an answer to "Find closed form for ∑n=1∞(−1)nn4Hn2n"

absolutestylehc

Answered question

2022-01-24

Find closed form for

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} n^{4} H_{n}}{2^{n}}

goleuedigdp

Beginner2022-01-25Added 7 answers

Recalling the generating function of the harmonic numbers

\sum_{n = 1}^{\infty} H_{n} x^{n} = \frac{\ln (1 - x)}{x - 1} \Rightarrow {(x D)}^{4} \sum_{n = 1}^{\infty} H_{n} x^{n}

= {(x D)}^{4} \frac{\ln (1 - x)}{x - 1}

\Rightarrow \sum_{n = 1}^{\infty} H_{\cap}^{4} x^{n}

= \frac{x (1 + 11 x + 11 x^{2} + x^{3}) \ln (1 - x)}{{(- 1 + x)}^{5}} + \frac{x (- 1 - 27 x^{2} - 18 x - 4 x^{3})}{(- 1 + x^{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 (\frac{3}{2})

waijazar1

Beginner2022-01-26Added 13 answers

\sum_{n = 1}^{\infty} \frac{(- 1)^{n} n^{4} H_{n}}{2^{n}} = \frac{28}{243} + \frac{10}{81} \log (\frac{2}{3})

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}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?