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

we have : $$\int^{\frac{\pi}{2}}_0\cos^{2n+2}(x)dx=\int^{\frac{\pi}{2}}_0\sin^{2n+2}(x)dx=\frac{(2n+1)!!}{(2n+2)!!}$$ therfore: $$S=\int^{\frac{\pi}{2}}_0\int^{\frac{\pi}{2}}_0\sum_{n=0}^{\infty} \frac{\cos^{2n+2}(x)\sin^{2n+2}(y)}{(2n+4)^2}dxdy$$

let $t:=\cos^2(x)\sin^2(y)$

therfore:

$$S=\int^{\frac{\pi}{2}}_0\int^{\frac{\pi}{2}}_0\sum_{n=0}^{\infty} \frac{t^{n+1}}{4(n+2)^2}dxdy$$

How to evaluate this :$$\sum_{n=0}^{\infty} \frac{t^{n+1}}{4(n+2)^2}$$

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  • $\begingroup$ 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. $\endgroup$
    – openspace
    Commented Jun 20 at 12:55
  • $\begingroup$ $Li_2 (t) = \sum_{n=1}^\infty \frac{t^n}{n^2}$. 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 $- \int^t_0 \dfrac{\ln (1-t)}{t} dt = Li_2 (t)$. $\endgroup$
    – Dave77
    Commented Jun 20 at 14:51

1 Answer 1

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$$\sum_{n=0}^{\infty} \left({\frac{(2n+1)!!}{(2n+2)!!}}\right)^2\frac{x^n}{(2n+4)^2}=\frac 1 \pi\sum_{n=0}^{\infty}\frac{\Gamma \left(n+\frac{3}{2}\right)^2}{\Gamma (n+2)^2}\,\frac{x^n}{(2n+4)^2}$$

As usual with gamma function in denominator, the result is a Gaussian hypergeometric function

$$\frac 1 {x^2} \,\, _2F_1\left(-\frac{1}{2},-\frac{1}{2};1;x\right)-\frac{x+4}{4 x^2}$$

which is also

$$\frac{4 \left(\frac{2 (x-1) }{\pi }K(x)+\frac{4 }{\pi }E(x)\right)-(x+4)}{4 x^2}$$ where appear elliptic integrals.

Make $x=1$ to get a very nice result.

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