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. 
 A: Good so far, to finish the proof notice that,
$$\frac{1}{1-z} = \sum_{i=0}^{\infty} z^i $$
Differentiating,
$$\frac{1}{(1-z)^2} = \sum_{i=1}^{\infty} i z^{i-1}$$
Multiply by $z$ then differentiate again,
$$z\frac{d}{dz} \frac{z}{(1-z)^2} = \sum_{i=1}^\infty i^2 z^{i} $$
So we have that,
$$\frac{z(z+1)}{(1-z)^3} = \sum_{k=1}^\infty k^2 z^k $$
Put in $z = x^2$ to obtain,
$$\sum_{k=1}^\infty k^2x^{2k} = \frac{x^2(x^2+1)}{(1-x^2)^3}$$
The most brute force way to calculate the integral after that is to substitute in $x = \sin \theta$, expand everything and separate and calculate all the integrals separately. 
A: I'm doing the final integration using symchdmath's result!
$$\frac1{3!}\int_0^1\frac{x^{-2}(1-x)^3x^2(1+x^2)}{(1-x^2)^3}dx=\frac1{3!}\int_0^1\frac{(1+x^2)}{(1+x)^3}dx=\frac1{3!}\int_0^1\frac{(1+x)^2-2(x+1)+2}{(1+x)^3}dx$$
which when separated into individual terms becomes
$$\frac1{3!}\Bigg[\int_0^1\frac1{1+x}dx-2\int_0^1\frac1{(1+x)^2}dx+2\int_0^1\frac1{(1+x)^3}dx\Bigg]=\frac1{3!}\bigg(ln2-1+\frac34\bigg)=\frac{4ln2-1}{24}$$
A: 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*}
A: Too long for a comment.
In the same spirit as @Tolaso's answer, using harmonic numbers in place of the digamma functions (just to have another set of notations), we can have a quite good approximation of the partial sums.
$$S_n= \sum_{k=0}^{n} \left ( \frac{1}{24\cdot 2\left ( k-\frac{1}{2} \right )}+ \frac{1}{8 \cdot 2 \left ( k + \frac{1}{2} \right )} - \frac{1}{12\left ( k+1 \right )} \right )=\frac{1}{12} \left(H_{n-\frac{1}{2}}-H_{n+1}+\frac{n-1}{2 n+1}+2\log (2)\right) $$ Using the asymptotics of the harmonic and Taylor series for the fraction, we should get
$$S_n=\frac{4 \log (2) -1}{24}-\frac{1}{16 n}+\frac{1}{16
   n^2}+O\left(\frac{1}{n^3}\right)$$ For $n=10$, the exact value is $S_{10}=\frac{95219407}{1396755360}\approx 0.06817$ while the above expansion would give $\frac{\log (2)}{6}-\frac{227}{4800}\approx 0.06823$.
A: Hint:
$$\sum t^k=f(t),$$
$$\sum kt^{k-1}=f'(t),$$
$$\sum kt^k=tf'(t),$$
$$\sum k^2t^{k-1}=(tf'(t)),$$
$$\sum k^2t^k=t(tf'(t))',$$
$$\sum k^2x^{2k}=x^2(x^2f'(x^2))'.$$
(Caution, the derivative is taken on $t$, not on $x$.)
A: An alternative way to evaluate the sum $\sum_{k=1}^\infty k^2 x^{2k}$ it is using finite calculus. Consider the indefinite sum $\sum (k^\underline 2+k) y^k\,\delta k$, then taking limits in this indefinite sum and substituting $y=x^2$ we recover your series, because $k^2=k^\underline 2+k$, where $k^\underline 2=k(k-1)$ is a falling factorial of order $2$.
In general it is easy to check that
$$\sum k^\underline m\,\delta k=\frac{k^\underline{m+1}}{m+1}+B,\quad\Delta_k k^\underline m=mk^\underline{m-1}\\ \sum c^k\,\delta k=\frac{c^k}{c-1}+B,\quad \Delta_k c^k=(c-1) c^k\tag1$$
where $B$ represent and arbitrary periodic function of period one and $c$ some arbitrary constant. Also we have the tool of summation by parts, that can be written as
$$\sum f(k)\Delta_k g(k)\, \delta k=f(k)g(k)-\sum g(k+1)\Delta_k f(k)\,\delta k\tag2$$
Thus, coming back to our sum, we have that
$$\sum (k^\underline 2+k)y^k\,\delta k=\frac1{y-1}y^k(k^\underline 2+k)-\frac1{y-1}\sum y^{k+1}(2k+1)\,\delta k\\
=\frac1{y-1}\left(y^k(k^\underline 2+k)-\left(\frac{y^{k+1}(2k+1)}{y-1}-\frac1{y-1}\sum y^{k+2}2\,\delta k\right)\right)\\
=\frac{y^k(k^\underline 2+k)}{y-1}-\frac{y^{k+1}(2k+1)}{(y-1)^2}+\frac{y^{k+2}2}{(y-1)^3}$$
where we applied twice summation by parts and the identities stated in $(1)$. Now, considering $|y|<1$, taking limits above we find that
$$\sum_{k=1}^\infty k^2 y^k=\frac{y}{1-y}+\frac{3y^2}{(1-y)^2}+\frac{2y^3}{(1-y)^3}=-y\frac{1+y}{(1-y)^3}$$
Then you can substitute back $y=x^2$ to find the final expression.
