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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, $$S =sum^{infty}_{k=1}frac{k^2cdot (2k-2)!}{(2k+2)!}=frac1{3!}sum^{infty}_{k=0}frac{(k+1)^2cdot(2k)!cdot 3!}{(2k+3+1)!}$$

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 $$S=sum^{infty}_{k=0}(k+1)^2int^{1}_{0}(x)^{2k}(1-x)^3dx$$

$$S=int^{1}_{0}x^{-2}(1-x)^3sum^{infty}_{k=1}(kx^k)^2dx$$

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.

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,

$$zfrac{d}{dz} frac{z}{(1-z)^2} = sum_{i=1}^infty i^2 z^{i} $$

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.

I'm doing the final integration using symchdmath's result!
$$frac1{3!}int_0^1frac{x^{-2}(1-x)^3x^2(1+x^2)}{(1-x^2)^3}dx=frac1{3!}int_0^1frac{(1+x^2)}{(1+x)^3}dx=frac1{3!}int_0^1frac{(1+x)^2-2(x+1)+2}{(1+x)^3}dx$$
which when separated into individual terms becomes
$$frac1{3!}Bigg[int_0^1frac1{1+x}dx-2int_0^1frac1{(1+x)^2}dx+2int_0^1frac1{(1+x)^3}dxBigg]=frac1{3!}bigg(ln2-1+frac34bigg)=frac{4ln2-1}{24}$$

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.
$$S_n= sum_{k=0}^{n} left ( frac{1}{24cdot 2left ( k-frac{1}{2} right )}+ frac{1}{8 cdot 2 left ( k + frac{1}{2} right )} - frac{1}{12left ( k+1 right )} right )=frac{1}{12} left(H_{n-frac{1}{2}}-H_{n+1}+frac{n-1}{2 n+1}+2log (2)right) $$ Using the asymptotics of the harmonic and Taylor series for the fraction, we should get
$$S_n=frac{4 log (2) -1}{24}-frac{1}{16 n}+frac{1}{16
n^2}+Oleft(frac{1}{n^3}right)$$ For $n=10$, the exact value is $S_{10}=frac{95219407}{1396755360}approx 0.06817$ while the above expansion would give $frac{log (2)}{6}-frac{227}{4800}approx 0.06823$.

Applying partial fractions along with the digamma function, we get that:

begin{align*}
sum_{k=1}^{infty} frac{k}{2left ( 2k-1 right )left ( 2k+1 right )left ( 2k+2 right )} &= sum_{k=1}^{infty} left ( frac{1}{24left ( 2k-1 right )}+ frac{1}{8left ( 2k+1 right )} -frac{1}{12left ( k+1 right )} right ) \
&=sum_{k=0}^{infty} left ( frac{1}{24left ( 2k-1 right )}+ frac{1}{8left ( 2k+1 right )} -frac{1}{12left ( k+1 right )} right ) \
&= sum_{k=0}^{infty} left ( frac{1}{24cdot 2left ( k-frac{1}{2} right )}+ frac{1}{8 cdot 2 left ( k + frac{1}{2} right )} - frac{1}{12left ( 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 -2log 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*}

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

$$sum k^underline m,delta k=frac{k^underline{m+1}}{m+1}+B,quadDelta_k k^underline m=mk^underline{m-1}\ sum c^k,delta k=frac{c^k}{c-1}+B,quad Delta_k c^k=(c-1) c^ktag1$$

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

$$sum f(k)Delta_k g(k), delta k=f(k)g(k)-sum g(k+1)Delta_k f(k),delta ktag2$$

Thus, coming back to our sum, we have that

$$sum (k^underline 2+k)y^k,delta k=frac1{y-1}y^k(k^underline 2+k)-frac1{y-1}sum y^{k+1}(2k+1),delta k\
=frac1{y-1}left(y^k(k^underline 2+k)-left(frac{y^{k+1}(2k+1)}{y-1}-frac1{y-1}sum y^{k+2}2,delta kright)right)\
=frac{y^k(k^underline 2+k)}{y-1}-frac{y^{k+1}(2k+1)}{(y-1)^2}+frac{y^{k+2}2}{(y-1)^3}$$

where we applied twice summation by parts and the identities stated in $(1)$. Now, considering $|y|<1$, taking limits above we find that

$$sum_{k=1}^infty k^2 y^k=frac{y}{1-y}+frac{3y^2}{(1-y)^2}+frac{2y^3}{(1-y)^3}=-yfrac{1+y}{(1-y)^3}$$

Then you can substitute back $y=x^2$ to find the final expression.

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 $x$.)