Krdytkyu

Finding sum of series $$displaystyle sum^{infty}_{k=1}frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$

Try: Let $$S = displaystyle sum^{infty}_{k=1}frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$

$$S = displaystyle sum^{infty}_{k=1}frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$

So $$S =sum^{infty}_{k=1}frac{k^2cdot (2k)!}{(2k+2)!}=sum^{infty}_{k=1}frac{k^2cdot(2k)!cdot 1!}{(2k+1+1)!}$$

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

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

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 to lengthy.

please explain 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.

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*}

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 $frac{log (2)}{6}-frac{227}{4800}approx 0.06823$.

Hint:

$$sum t^k=f(t),$$

$$sum kt^{k-1}=f'(t),$$

$$sum kt^k=tf'(t),$$

$$sum k^2t^{k-1}=(tf'(t)),$$

$$sum k^2t^k=t(tf'(t))',$$

$$sum k^2x^{2k}=x^2(x^2f'(x^2))'.$$