Closed form expression for $\sum\limits_{k=0}^{\infty}\frac{k4^k}{n+k}{n+k\choose n-k}{n\choose k}{2n\choose 2k}^{-1}$

Question

I'm trying to find a closed form expression (that doesn't involve an indefinite summation) for the following combinatorial sum : $$\sum\limits_{k=0}^{n}\left[\frac{k4^k}{n+k}{n+k\choose n-k}\frac{\displaystyle{n\choose k}}{\displaystyle{2n\choose 2k}}\right]$$ I don't know wether or not there is one in the first place. The context in which this sum arised isn't relevant and won't provide any help.

I don't have any idea as of where to even start. Any suggestion ?

Answer 1

Using Wolfgang Kais' rearrangement,

$$ \frac{4^n}{\binom{2n}{n}}\sum_{k=0}^{n-1}\frac{1}{4^k}\binom{2k}{k}\binom{n+(n-1-k)}{n} $$ is a convolution which equals $$ \frac{4^n}{\binom{2n}{n}}[x^{n-1}]\sum_{m\geq 0}\frac{1}{4^m}\binom{2m}{m}x^m \sum_{m\geq 0}\binom{n+m}{n}x^m $$ or $$ \frac{4^n}{\binom{2n}{n}} [x^{n-1}]\frac{1}{\sqrt{1-x}}\cdot\frac{1}{(1-x)^{n+1}}=\frac{4^n}{\binom{2n}{n}}[x^{n-1}](1-x)^{-(n+3/2)}=\frac{(-1)^{n+1}4^n\binom{-n-3/2}{n-1}}{\binom{2n}{n}} $$ which simplifies into $$\frac{2 \sqrt{\pi } n \,\Gamma \left(2 n+\frac{1}{2}\right)}{(2 n+1)\, \Gamma \left(n+\frac{1}{2}\right)^2}=\color{red}{\frac{2n}{2n+1}\cdot\frac{\binom{4n}{2n}}{\binom{2n}{n}}}.$$

Answer 2

You can always look for an expression in terms of the hypergeometric function, which has a lot of nice properties.

Indeed, Wolframalpha says your sum can be written like this: $$\frac{2n}{2n-1}\ _2F_1(1-n,n+1;3/2-n;1)$$ The website also offers a small list of values for different $n$. This list indicates divergence of the sum.

But I don't see how this will be helpful, unless there is more information about what you want to do with it.

Answer 3

$$S_n=\sum\limits_{k=0}^{n}\left[\frac{k4^k}{n+k}{n+k\choose n-k}\frac{\displaystyle{n\choose k}}{\displaystyle{2n\choose 2k}}\right]=-\frac{n}{\sqrt \pi}\frac{ \Gamma \left(-n-\frac{1}{2} \right) \Gamma \left(-n+\frac{1}{2} \right)}{ \Gamma \left(\frac{1}{2}-2n\right)}$$

For large values of $n$, using Stirling-like approximations, $$\log(S_n)=2 n \log (2)-\frac{\log (2)}{2}-\frac{7}{16 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$S_n \sim 2^{2 n-\frac{1}{2}}\exp\left(-\frac{7}{16 n}+\frac{1}{8 n^2} \right)$$ which seems to be quite good even for small $n$ $$\left( \begin{array}{ccc} n & \text{aproximation} & \text{exact} \\ 1 & 2.06932 & 2.00000 \\ 2 & 9.37939 & 9.33333 \\ 3 & 39.6609 & 39.6000 \\ 4 & 163.537 & 163.429 \\ 5 & 666.739 & 666.508 \\ 6 & 2702.00 & 2701.45 \\ 7 & 10911.1 & 10909.7 \\ 8 & 43960.5 & 43956.7 \\ 9 & 176841. & 176830. \\ 10 & 710604. & 710572. \end{array} \right)$$