Finding a closed form for $\sum_{k=1}^\infty\sum_{n=k}^\infty\left(\frac{(-1)^k}{k^3\binom{n+k}{k}\binom{n}{k}}(\frac1{n^2}-\frac1{(n+1)^2})\right)$

Question

Consider the sum $$\sum_{n=k}^{N} \frac{1}{\binom{n+k}{k}\binom{n}{k}}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) $$ Using the above or otherwise I need a closed form for $$\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\left(\frac{(-1)^k}{k^3 \binom{n+k}{k}\binom{n}{k}}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\right) $$

I tried as follows: Denote $$S_{N,k}:=\sum_{n=k}^{N} \frac{1}{\binom{n+k}{k}\binom{n}{k}}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) $$

On writing the formula for $n\choose k$ we get $$S_{N,k}:=\sum_{n=k}^{N} \frac{(n-k)!}{(n+k)!}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) $$ Now $$\frac{(n-k)!}{(n+k)!}=\frac{(n-k)(n-k-1)!}{(n+k)!} $$ Using partial fractions we have $$\frac{(n-k-1)!}{(n+k)!}=\frac{1}{(n-k)...(n+k)}=\frac{1}{(2k)!(n-k)}-\frac{1}{(2k-1)!1!(n-k+1)}+\frac{1}{(2k-2)!2!(n-k+2)}-... $$ So we have $$\frac{(n-k-1)!}{(n+k)!}=\frac{1}{(2k)!}\sum_{m=0}^{2k}\frac{(-1)^m \binom{2k}{m}}{n-k+m}$$ Plugging the above partial fraction decomposition in $S_{N,k}$ we get $$S_{N,k}=\frac{1}{(2k)!}\sum_{n=k}^{N} \left(\sum_{m=0}^{2k}\frac{(-1)^m \binom{2k}{m} (n-k)}{n-k+m}\right)\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) $$ $$S_{N,k}=\frac{1}{(2k)!}\sum_{n=k}^{N} \left(\sum_{m=0}^{2k}(-1)^m \binom{2k}{m} \left(1-\frac{m}{n-k+m}\right)\right)\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) $$ Now since $$\sum_{m=0}^{2k}(-1)^m \binom{2k}{m}=0$$ So we get $$S_{N,k}=-\frac{1}{(2k)!}\sum_{n=k}^{N} \left(\sum_{m=1}^{2k} \frac{(-1)^m\ m \ \binom{2k}{m}}{n-k+m}\right)\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) $$ After this I am unable to simplify further. Please help me to find the sum.

Edit Using the above or otherwise I need a closed form for $$\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\left(\frac{(-1)^k}{k^3 \binom{n+k}{k}\binom{n}{k}}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\right) $$

Answer 1

This is not an answer - just observations

Let

$$a_k=\sum_{n=k}^{\infty}\frac{(-1)^k}{k^3 \binom{n+k}{k}\binom{n}{k}}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)$$ Computing the very first terms $$a_k=b_k - \frac{2\, \zeta(3)}{k^2}$$ where the first $b_k$ form the sequence $$\left\{2,\frac{29}{48},\frac{5191}{19440},\frac{72703}{483840}, \frac{7270009}{75600000},\frac{47982293}{718502400},\frac{305647 00141}{622961539200}\right\}$$ $$S_p=\sum_{k=1}^p a_k=-2\, \zeta (3)\, H_p^{(2)}+\sum_{k=1}^p b_k$$

The first terms of the summation are $$\left\{2,\frac{125}{48},\frac{6977}{2430},\frac{13157111}{4354560 },\frac{8484914699}{2721600000},\frac{71499992923}{22453200000 },\frac{3218912736881}{995499618750}\right\}$$ which looks very much like an hyperbolic function.

Using the empirical $$\sum_{k=1}^p b_k=\alpha -\frac \beta{p+\gamma}$$ we have with $R^2=0.999999986$

$$\begin{array}{|llll} \hline \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \alpha & 3.55728 & 0.00048 & \{3.55612,3.55845\} \\ \beta & 2.45307 & 0.00347 & \{2.44458,2.46156\} \\ \gamma & 0.57511 & 0.00192 & \{0.57042,0.57979\} \\ \end{array}$$

So, the infinite sum $$\sum_{k=1}^\infty a_k \sim \alpha - \frac{\pi ^2 }{3}\zeta (3)$$ for a suitable $\alpha$.

Answer 2

The transformation of $S_{N,k}$ needs to be revised somewhat, since we have \begin{align*} \binom{n+k}{k}\binom{n}{k}&=\frac{(n+k)!}{n!k!}\,\frac{n!}{k!(n-k)!}\\ &=\frac{(2k)!}{k!k!}\,\frac{(n+k)!}{(2k)!(n-k)!}=\binom{2k}{k}\binom{n+k}{n-k}\tag{1} \end{align*}

We obtain from (1) \begin{align*} \color{blue}{S_{N,k}}&\color{blue}{=\sum_{n=k}^N\binom{n+k}{k}^{-1}\binom{n}{k}^{-1} \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)}\\ &=\binom{2k}{k}^{-1}\sum_{n=k}^{N}\binom{n+k}{n-k}^{-1}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{2}\\ &=\binom{2k}{k}^{-1}\sum_{n=k}^{N}(n+k+1)\int_{0}^1z^{n-k}(1-z)^{2k}\,dz \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{3}\\ &=\binom{2k}{k}^{-1}\int_0^1\frac{(1-z)^{2k}}{z^k} \color{blue}{\sum_{n=k}^N(n+k+1)\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)z^n}\,dz\\ \end{align*}

Regrettably, I do not see a nice simplification of the inner sum. We can rewrite it using the Lerch transcendent. We have for instance \begin{align*} \sum_{n=k}^N\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)z^n &=z^k\left(\Phi(z,2,k)-\Phi(z,2,k+1)\right)\\ &\qquad-z^{N+1}\left(\Phi(z,2,N+1)-\Phi(z,2,N+2)\right) \end{align*} but it's rather just another representation, not a simplification of the sum.

Comment:

• In (2) we apply the binomial identity (1).

• In (3) we write the reciprocal of a binomial coefficient using the beta function \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz \end{align*}