A closed form for $\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$

Question

Mathematica does give an analytic form for $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}.$$
The question here is: How to find a simpler closed form for this alternating summation by hand. The summation of the absolute seies has been discussed earlier in MSE: Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$

Answer 1

Since

$$ \frac{\binom nk}{k+1}=\frac{\binom{n+1}{k+1}}{n+1}\;, $$

we have

\begin{eqnarray} \sum_{k=0}^n\frac{(-1)^k\binom nk^2}{(k+1)^2} &=& \frac1{(n+1)^2}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}^2 \\ &=& \frac1{(n+1)^2}\left(1-\sum_{k=0}^{n+1}(-1)^k\binom{n+1}k^2\right)\;. \end{eqnarray}

With

$$ \sum_{k=0}^nq^k\binom nk^2=(1-q)^nP_n\left(\frac{1+q}{1-q}\right)\;, $$

where $P_n$ is the $n$-th Legendre polynomial, this is

$$ \frac1{(n+1)^2}\left(1-2^{n+1}P_{n+1}(0)\right)\;, $$

where

$$ P_l(0)= \begin{cases} \frac{(-1)^m}{4^m}\binom{2m}m&l=2m\\0&l=2m+1 \end{cases} $$

(see Wikipedia).

The factor $4^m$ cancels, so the result is

$$ \sum_{k=0}^n\frac{(-1)^k\binom nk^2}{(k+1)^2}= \begin{cases} \frac{1-(-1)^m\binom{2m}m}{(2m)^2}&n=2m-1\;,\\ \frac{1}{(2m+1)^2}&n=2m\;. \end{cases} $$