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

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.

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  • $\begingroup$ Easier: Use partial fractions on the summand first, then sum. $\endgroup$ – user10354138 Dec 16 '18 at 7:55
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    $\begingroup$ There's an error in your third line, it should be $S =\sum^{\infty}_{k=1}\frac{k^2\cdot (2k-2)!}{(2k+2)!}=1/3! \sum^{\infty}_{k=0}\frac{(k+1)^2\cdot(2k)!\cdot 3!}{(2k+3+1)!}$. Then you are looking for $S=1/3! \sum^{\infty}_{k=0}(k+1)^2\int^{1}_{0}x^{2k}(1-x)^3dx = 1/3! \int^{1}_{0}\left[\sum^{\infty}_{k=1}k^2x^{2k}\right] x^{-2}(1-x)^3dx$. You can then reuse the result of @symchdmath, and you're left with $1/6 \int_0^1 \frac{1+x^2}{(1+x)^3} dx$, which you can solve with partial fraction expansion. $\endgroup$ – yoann Dec 16 '18 at 10:09
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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.

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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 $$\frac1{3!}\Bigg[\int_0^1\frac1{1+x}dx-2\int_0^1\frac1{(1+x)^2}dx+2\int_0^1\frac1{(1+x)^3}dx\Bigg]=\frac1{3!}\bigg(ln2-1+\frac34\bigg)=\frac{4ln2-1}{24}$$

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

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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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  • $\begingroup$ This does not address the question, does it ? (I admit that the OP was unclear.) $\endgroup$ – Yves Daoust Dec 16 '18 at 8:41
  • $\begingroup$ Well it does answer how to evaluate the series , but not how to continue with the OP's approach. $\endgroup$ – Tolaso Dec 16 '18 at 11:28
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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.

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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 $x$.)

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