Calculate $\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)$ 
Denote $a_n=\frac{1}{3}\cdot\frac{2}{5}\cdots\frac{n}{2n+1}\cdot\frac{1}{n+1}$.
Please prove
$$\sum_{n=1}^{\infty}a_n=\frac{\pi^2}{8}-1$$

This is the answer given by my friend. He also used the integral $\int_{-\pi}^{\pi}\cos^kx\,dx$ along with much calculation in which the answer came out almost accidentally.
I want to see other ways to deal with it. Feel free to use the integral above. Thanks in advance.
Edit:This may be a little long
(I)$\int_{0}^{\pi}\sin^{2n+1}xdx=\frac{2^n\dot n!}{(2n+1)!!}$
(II)$\sum_{n=1}^\infty \frac{x^{n+1}}{n+1}=-x-ln(1-x)$
(III)$-\int_{-1}^1\frac{ln(\frac{1+x^2}{2})}{1-x^2}dx=-\int_{-1}^1\frac{ln(1-\frac{1-x^2}{2})}{1-x^2}dx=-\int_{-1}^1\frac{ln(1-t(1-x^2))}{1-x^2}\big|_{t=0}^{1/2}dx=\int_{-1}^1\int_{0}^{1/2}\frac{dtdx}{1-t(1-x^2)}=\int_{0}^{1/2}\int_{-1}^1\frac{dxdt}{1-t(1-x^2)}=2\int_0^{1/2}\frac{\arctan \sqrt{\frac{t}{1-t}}}{\sqrt{t(1-t)}}dt=4\int_0^{1/2}\arctan \sqrt{\frac{t}{1-t}}d\arctan\sqrt{\frac{t}{1-t}}=4\cdot\frac{1}{2}\cdot(\frac{\pi}{4})^2=\frac{\pi^2}{8}$
(IV)$\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)=\sum_{n=1}^{\infty}\frac{n!}{(2n+1)!!}\frac{1}{n+1}=\sum_{n=1}^{\infty}\frac{\int_{0}^{\pi}\sin^{2n+1}xdx}{2^{n+1}}\frac{1}{n+1}=\int_{0}^{\pi}\sum_{n=1}^{\infty}\frac{\sin^{2n+1}x}{2^{n+1}}\frac{dx}{n+1}=\int_{0}^{\pi}\frac{1}{\sin x}\sum_{n=1}^{\infty}(\frac{\sin^{2}x}{2})^{n+1}\frac{dx}{n+1}=\int_0^{\pi}\frac{-\frac{\sin^{2}x}{2}-ln(1-\frac{\sin^{2}x}{2})}{\sin x}dx=-1+\int_0^{\pi}\frac{-ln(1-\frac{\sin^{2}x}{2})}{\sin x}dx=-1+\int_0^{\pi}\frac{ln(1-\frac{\sin^{2}x}{2})}{\sin^2 x}d\cos x=-1+\int_0^{\pi}\frac{ln(\frac{1+\cos^{2}x}{2})}{1-\cos^2 x}d\cos{x}=-1+\frac{\pi^2}{8}$
(Sorry for ugly typing style.)
 A: Notice we can decompose $a_n$ as a product of two factors
$$a_n = \frac{1}{n+1}\prod_{k=1}^n \frac{k}{2k+1} = \frac{1}{n+1}\frac{n!}{(2n+1)!!}$$
and both factors can be represented as integrals.
$$\begin{align}
\frac{1}{n+1} &= \int_0^1 x^{n} dx\\
\frac{n!}{(2n+1)!!} &= \frac{2^n n!^2}{(2n+1)!} =  \frac{2^n\Gamma(n+1)^2}{\Gamma(2n+2)}  = \int_0^1 (2t(1-t))^n dt
\end{align}$$
Using these integral representations, we obtain
$$\mathcal{I} \stackrel{def}{=} \sum_{n=0}^\infty a_n = \int_0^1\int_0^1 \sum_{n=0}^\infty (2xt(1-t))^n dt dx
= \int_0^1\int_0^1 \frac{1}{1-2xt(1-t)} dt dx$$
Change variable to $t = \frac{1+u}{2}$, the integral becomes
$$\begin{align}\mathcal{I}
&= \int_0^1\int_{-1}^1 \frac{1}{2 - x(1-u^2)} du dx
= \int_0^1 \int_0^1 \frac{2}{(2-x) + xu^2} du dx\\
&= \int_0^1 \frac{2}{\sqrt{x(2-x)}}\tan^{-1}\sqrt{\frac{x}{2-x}} dx\end{align}$$
Change variable to $y = 1-x$ and then to $\theta = \cos^{-1}y$, we get
$$\begin{align}\mathcal{I} 
&= \int_0^1 \frac{2}{\sqrt{1-y^2}}\tan^{-1}\sqrt{\frac{1-y}{1+y}} dy
= 2\int_0^{\pi/2}\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} d\theta\\
&= 2\int_0^{\pi/2}\tan^{-1}\left(\tan\frac{\theta}{2}\right) d\theta
= \int_0^{\pi/2}\theta d\theta
= \frac{\pi^2}{8}\end{align}$$
As a result, $$\sum_{n=1}^\infty a_n = \sum_{n=0}^\infty a_n - a_0 = \mathcal{I} - 1 = \frac{\pi^2}{8} - 1$$
A: Using the Factorial aswell as the Double Factorial we may rewrite $a_n$ as
$$a_n=\frac13\frac25\cdots\frac n{2n+1}\frac1{n+1}=\frac{n!}{(2n+1)!!}\frac1{n+1}$$
Expressing the utilized Double Factorial in terms of the Factorial this can be further reduced
$$a_n=\frac1{(2n+1)!!}\frac{n!}{n+1}=\frac{2^{n+1}(n+1)!}{(2n+2)!}\frac{n!}{n+1}=\frac{2^n}{(n+1)(2n+1)\binom{2n}n}$$
Now, the sum comes into play. Plugging our new gained expression within the summation followed by an index shift gives us the following
\begin{align*}
\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{2^n}{(n+1)(2n+1)\binom{2n}n}&=\sum_{n=2}^\infty\frac{2^{n-1}}{n(2n-1)\binom{2n-2}{n-1}}\\
&=\sum_{n=2}^\infty \frac{2^n}{(2n)(2n-1)\frac{(2n-2)!}{(n-1)!^2}}\\
&=\sum_{n=2}^\infty \frac{2^n}{n^2\frac{(2n)!}{n!^2}}\\
&=\sum_{n=2}^\infty\frac{2^n}{n^2\binom{2n}n}
\end{align*}
The last sum is of a truly remarkable structure, to be precise the famous series expansion of the squared inverse sine function! Applying this knowledge we finally obtain
$$\sum_{n=2}^\infty\frac{2^n}{n^2\binom{2n}n}=2\arcsin^2\left(\frac{\sqrt2}2\right)-1=\frac{\pi^2}8-1$$

$$\therefore~\sum_{n=1}^\infty a_n~=~\frac{\pi^2}8-1$$

