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Proof of $\sum_{k=0}^{\infty}{\frac{1}{(k+1)(k+3)}}={\frac { \pi \csc \left(
\pi \right) -\sec \left( \pi \right) \csc ^{2}\left( \pi
\right) }{\pi }}$

How to prove $\sum_{k=0}^{\infty}{\frac{1}{(k+1)(k+3)}}={\frac {
\pi \csc \left( \pi \right) -\sec \left( \pi \right) \csc
^{2}\left( \pi \right) }{\pi }}$?
I tried to make substitutions $t=k+1$ and $s=k+3$ but it resulted in a messy series and I am not sure how to procede. Any help would be appreciated.

A:

Hint: $$\frac1{k+1}=\int_0^\infty t^{k}\,e^{ -t}\,dt,\qquad\frac1{k+3}=\int_0^\infty t^{k+2}\,e^{ -t}\,dt$$
$$\frac1{(k+1)(k+3)}=\int_0^\infty t^k\,e^{ -t}\int_0^t t^{k+2}\,e^{ -t}\,dt\,dt=\int_0^\infty \frac{t^k}2\,dt$$
Can you finish from here?

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