A closed form for $\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$ 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 $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$
 A: 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
$$
\sum_{k=0}^n\frac{(-1)^k\binom nk^2}{(k+1)^2}=
\begin{cases}
\frac{1-(-1)^m\binom{2m}m}{(2m)^2}&n=2m-1\;,\\
\frac{1}{(2m+1)^2}&n=2m\;.
\end{cases}
$$
A: Use Binomial identity:
$$
(1+t)^n=\sum_{k=0}^{n} {n \choose k}t^n.
\tag{1}
$$
Integration of $(1)$ from $t=0$ to $t=x$ gives
$$
\frac{(1+x)^{n+1}-1}{n+1}= \sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}.\tag{2}
$$
We can change $x$ to $-1/x$ in $(2)$ to get
$$
\frac{(1-1/x)^{n+1}-1}{n+1}= \sum_{k=0}^n (-1)^k {n \choose k}\frac{x^{-k-1}}{k+1}
.
\tag{3}
$$
Multiplying $(2)$ and $(3)$ and collecting terms free of $x$ on RHS, we get
$$
\frac{x^{-n-1}\big[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}\big]}{(n+1)^2}=\sum_{k=0}^{n}\frac{ (-1)^k {n \choose k}^2}{(k+1)^2} x^0+\dots
$$
Equating the coefficients yields
$$S_n = \sum_{j=0}^{n} \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$$
$$S_N=[x^{n+1}]~\frac{[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}]}{(n+1)^2} $$
$$S_n=\frac{-(-1)^{(n+1)/2}{n+1 \choose (n+1)/2}+1}{(n+1)^2}, \text{when $n$ is odd}$$
$$S_n= \frac{1}{(n+1)^2}, \text{when $n$ is even} $$
