Timeline for I need Help to evaluate :$\sum_{n=0}^{\infty} \left({\frac{(2n+1)!!}{(2n+2)!!}}\right)^2\frac{1}{(2n+4)^2}$

6 events

when toggle format	what		by	license	comment
Jun 20 at 15:06	vote	accept	Mostafa
Jun 20 at 15:06	vote	accept	Mostafa
Jun 20 at 15:06
Jun 20 at 14:51	comment	added	Dave77		$Li_2 (t) = \sum_{n=1}^\infty \frac{t^n}{n^2}$. You can obtain an integral expression: $- \ln (1-t) = \sum_{n=1}^\infty \frac{t^n}{n}$. So that $- \int^t_0 \dfrac{\ln (1-t)}{t} dt = \sum_{n=1}^\infty \frac{t^n}{n^2}$ Then $- \int^t_0 \dfrac{\ln (1-t)}{t} dt = Li_2 (t)$.
Jun 20 at 14:01	answer	added	Claude Leibovici		timeline score: 1
Jun 20 at 12:55	comment	added	openspace		Why do you think the given sum would have an explicit form? The general way of solving such series: let $S(t) = \sum_{n \ge 0} \frac{t^{n+2}}{4(n+2)}$, hence $S'(t) = \sum_{n \ge 0} \frac{t^{n+1}}{4(n+2)} \Rightarrow (S'(t)t)' = \sum_{n \ge 0} t^{n+1}$. Therefore, you'll obtain an ODE to solve.
Jun 20 at 12:27	history	asked	Mostafa	CC BY-SA 4.0