A Closed form for the $\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{(25k^2+25k+4)(n-k+1)^3}$

Question

I'm looking for a closed form for this sequence,

$$\sum_{n=1}^{\infty}\left(\sum_{k=1}^{n}\frac{1}{(25k^2+25k+4)(n-k+1)^3} \right)$$

I applied convergence test. The series converges.I want to know if the series is expressed with any mathematical constant. How can we do that?

Answer 1

Change the order of summation, so it's $\sum_{k=1}^\infty \sum_{n=k}^\infty$. Then I get $$ {\frac {\zeta \left( 3 \right) \left( 4\,\pi\,\cot \left( \pi/5 \right) -15 \right) }{60}} $$ You could also write $$\cot(\pi/5) = \frac{\sqrt{2}}{20} (5 + \sqrt{5})^{3/2}$$

Answer 2

Proceeding in my usual naive way,

$\begin{array}\\ S &=\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{(25k^2+25k+4)(n-k+1)^3}\\ &=\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{(25k^2+25k+4)(n-k+1)^3} \\ &=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\frac{1}{(25k^2+25k+4)(n-k+1)^3} \\ &=\sum_{k=1}^{\infty}\frac1{(25k^2+25k+4)}\sum_{n=k}^{\infty}\frac{1}{(n-k+1)^3} \\ &=\sum_{k=1}^{\infty}\frac1{(25k^2+25k+4)}\sum_{n=1}^{\infty}\frac{1}{n^3} \\ &=\zeta(3)\sum_{k=1}^{\infty}\frac1{(25k^2+25k+4)}\\ &=\zeta(3)\sum_{k=1}^{\infty}\frac1{(5k+1)(5k+4)}\\ &=\zeta(3)\sum_{k=1}^{\infty}\frac13\left(\frac1{5k+1}-\frac1{5k+4}\right)\\ &=\frac{\zeta(3)}{3}\lim_{m \to \infty} \sum_{k=1}^{m}\left(\frac1{5k+1}-\frac1{5k+4}\right)\\ &=\frac{\zeta(3)}{15}\lim_{m \to \infty} \left(\sum_{k=1}^{m}\frac1{k+1/5}-\sum_{k=1}^{m}\frac1{k+4/5}\right)\\ &=-\frac{\zeta(3)}{15}\lim_{m \to \infty} \left(\sum_{k=1}^{m}(\frac1{k}-\frac1{k+1/5})-\sum_{k=1}^{m}(\frac1{k}-\frac1{k+4/5})\right)\\ &=-\frac{\zeta(3)}{15}(\psi(6/5)-\psi(9/5))\\ &=\frac{\zeta(3)}{15}(\psi(9/5)-\psi(6/5))\\ \end{array} $

where $\psi(x)$ is the digamma function (https://en.wikipedia.org/wiki/Digamma_function).

Note: Wolfy says that $\sum_{k=1}^{\infty}\frac1{(5k+1)(5k+4)} = \frac{\pi}{15}\sqrt{1 + \frac{2}{\sqrt{5}}} - \frac1{4} $.