<s>[INST] Evaluating $\sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$ 
Finding sum of series $$\displaystyle \sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$

Try: Let $$S = \displaystyle \sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$
So, $$S =\sum^{\infty}_{k=1}\frac{k^2\cdot (2k-2)!}{(2k+2)!}=\frac1{3!}\sum^{\infty}_{k=0}\frac{(k+1)^2\cdot(2k)!\cdot 3!}{(2k+3+1)!}$$
with the help of identity $$B(m,n) = \int^{1}_{0}x^m(1-x)^ndx = \frac{\Gamma (m+1)\Gamma(n+1)}{\Gamma(m+n+2)}$$
$$B(m,n) = \frac{\Gamma (m+1)\Gamma(n+1)}{\Gamma(m+n+2)}=\frac{m!\cdot n!}{(m+n+1)!}$$
So $$S=\sum^{\infty}_{k=0}(k+1)^2\int^{1}_{0}(x)^{2k}(1-x)^3dx$$
$$S=\int^{1}_{0}x^{-2}(1-x)^3\sum^{\infty}_{k=1}(kx^k)^2dx$$
Can someone explain me how to calculate $\displaystyle \sum^{\infty}_{k=1}k^2x^{2k}$ in some short way . although I am trying to solve it but it is too lengthy. 
Please explain to me ,thanks. 
 [/INST] Applying partial fractions along with the digamma function, we get that:
\begin{align*}
\sum_{k=1}^{\infty} \frac{k}{2\left ( 2k-1 \right )\left ( 2k+1 \right )\left ( 2k+2 \right )} &= \sum_{k=1}^{\infty} \left ( \frac{1}{24\left ( 2k-1 \right )}+ \frac{1}{8\left ( 2k+1 \right )} -\frac{1}{12\left ( k+1 \right )} \right ) \\ 
 &=\sum_{k=0}^{\infty}  \left ( \frac{1}{24\left ( 2k-1 \right )}+ \frac{1}{8\left ( 2k+1 \right )} -\frac{1}{12\left ( k+1 \right )} \right ) \\ 
 &= \sum_{k=0}^{\infty} \left ( \frac{1}{24\cdot 2\left ( k-\frac{1}{2} \right )}+ \frac{1}{8 \cdot 2 \left ( k + \frac{1}{2} \right )} - \frac{1}{12\left ( k+1 \right )} \right )\\ 
 &= -\frac{1}{48} \psi^{(0)} \left ( -\frac{1}{2} \right ) - \frac{1}{16} \psi^{(0)} \left ( \frac{1}{2} \right ) + \frac{1}{12} \psi^{(0)} (1) \\ 
 &=-\frac{1}{48} \left ( 2-\gamma -2\log 2 \right ) - \frac{1}{16} \left ( -\gamma -2 \log 2  \right ) - \frac{\gamma}{12}  \\ 
 &= \frac{1}{24} \left ( \log 16 -1 \right ) \\
 & =\frac{4 \log 2 -1}{24}
\end{align*}
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