TITLE: Closed form for a binomial series
QUESTION [7 upvotes]: I am wondering if any knows how to compute a closed form for the following two series. 

$$\sum_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m}$$
$$\sum_{m=1}^{n}\frac{(-1)^m}{m^4}\binom{2n}{n+m}$$

Mathematica gave a result in terms of the Hypergeometric function, which I don't understand. I was wondering if anyone can illustrate how to express these two series in a nice closed form, without using the Hypergeomeyric function. I was thinking along the line using a combinatorial identity or method, but I having problems figuring it out.

REPLY [8 votes]: Let
$$
g(n,m)=\frac1{\binom{2n}{n}}\sum_{j=1}^n\frac{(-1)^j}{j^m}\binom{2n}{n+j}\tag{1}
$$
then it can be shown that
$$
g(n,m)=g(n-1,m)+\frac{g(n,m-2)}{n^2}\tag{2}
$$
The proof of $(2)$ is below.
Since
$$
\sum_{j=0}^k(-1)^j\binom{n}{j}=(-1)^k\binom{n-1}{k}\tag{3}
$$
we get
$$
g(n,0)=-\frac12\tag{4}
$$
then, using $(2)$ and $(4)$, we get
$$
g(n,2)=-\frac12H_n^{(2)}\tag{5}
$$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{j=1}^n\frac{(-1)^j}{j^2}\binom{2n}{n+j}=-\frac12H_n^{(2)}\binom{2n}{n}}\tag{6}
$$
where
$$
H_n^{(k)}=\sum_{j=1}^n\frac1{j^k}\tag{7}
$$

Using $(2)$ and $(5)$, we get that
$$
g(n,4)=-\frac14\left(\left(H_n^{(2)}\right)^2+H_n^{(4)}\right)\tag{8}
$$
therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{j=1}^n\frac{(-1)^j}{j^4}\binom{2n}{n+j}=-\frac14\left(\left(H_n^{(2)}\right)^2+H_n^{(4)}\right)\binom{2n}{n}}\tag{9}
$$

Proof of $\boldsymbol{(2)}$:
Note that
$$
\begin{align}
g(n,m)
&=\sum_{j=1}^n\frac{(-1)^j}{j^m}\frac{n!}{(n-j)!}\frac{n!}{(n+j)!}\\
&=\frac{(-1)^n}{n^m\binom{2n}{n}}+\sum_{j=1}^{n-1}\frac{(-1)^j}{j^m}\frac{(n-1)!}{(n-j-1)!}\frac{(n-1)!}{(n+j-1)!}\frac{n^2}{n^2-j^2}\\
g(n-1,m)
&=\hphantom{\frac{(-1)^n}{n^m\binom{2n}{n}}+}\sum_{j=1}^{n-1}\frac{(-1)^j}{j^m}\frac{(n-1)!}{(n-j-1)!}\frac{(n-1)!}{(n+j-1)!}\\
g(n,m)-g(n-1,m)
&=\frac{(-1)^n}{n^m\binom{2n}{n}}+\sum_{j=1}^{n-1}\frac{(-1)^j}{j^m}\frac{(n-1)!}{(n-j-1)!}\frac{(n-1)!}{(n+j-1)!}\frac{j^2}{n^2-j^2}\\
&=\frac{(-1)^n}{n^m\binom{2n}{n}}+\frac1{n^2}\sum_{j=1}^{n-1}\frac{(-1)^j}{j^{m-2}}\frac{n!}{(n-j)!}\frac{n!}{(n+j)!}\\
&=\frac1{n^2}\sum_{j=1}^n\frac{(-1)^j}{j^{m-2}}\frac{n!}{(n-j)!}\frac{n!}{(n+j)!}\\
&=\frac{g(n,m-2)}{n^2}
\end{align}
$$