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 $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$

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
$$S_n = \sum_{j=0}^{n} \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$$
$$S_N=[x^{n+1}]~\frac{[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}]}{(n+1)^2} $$
$$S_n=\frac{-(-1)^{(n+1)/2}{n+1 \choose (n+1)/2}+1}{(n+1)^2}, \text{when $n$ is odd}$$
$$S_n= \frac{1}{(n+1)^2}, \text{when $n$ is even} $$
