How to prove $\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \frac{1}{N+1} \sum\limits_{n=1}^{N+1} \frac{1}{n}$ In the process of proving a more complicated relation, I've come across the following equality that I'm having trouble proving:
$$
\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \frac{1}{N+1} \sum\limits_{n=1}^{N+1} \frac{1}{n}
$$
I was already able to prove the following similar equality:
$$
\sum\limits_{k=0}^N \frac{(-1)^k {N \choose k}}{k+1} = \frac{1}{N + 1}
$$
but I'm unsure how to proceed with the first one. I assume it has something to do with the fact that every term in the left hand side of the first equality is $\frac{1}{k+1}$ times a term in the left hand side of the second equality. Any help would be greatly appreciated.

Introduce the function
$$f(z) = (-1)^n \times n! \times \frac{1}{(1+z)^2}
\prod_{q=0}^n \frac{1}{z-q}.$$
Then we have for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z)
= (-1)^n \times n! \times \frac{1}{(1+k)^2}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^{n} \frac{1}{k-q}
\\ = (-1)^n \times n! \times \frac{1}{(1+k)^2}
\frac{1}{k!} (-1)^{n-k} \frac{1}{(n-k)!}
= {n\choose k} (-1)^k \frac{1}{(1+k)^2}.$$
Now as residues sum to zero we get
$$\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+k)^2}
+ \mathrm{Res}_{z=-1} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$
The residue at infinity is zero by inspection. We also have
$$\mathrm{Res}_{z=-1} f(z) =
 (-1)^n \times n! \times
\left.\left(\prod_{q=0}^n \frac{1}{z-q}\right)'\right|_{z=-1}
\\ =  (-1)^n \times n! \times
\left.\prod_{q=0}^n \frac{1}{z-q}
\sum_{q=0}^n \frac{1}{q-z} \right|_{z=-1}
\\ =  (-1)^n \times n! \times
\prod_{q=0}^n \frac{1}{-1-q}
\sum_{q=0}^n \frac{1}{q+1}
=  (-1)^n \times n! \times
\frac{(-1)^{n+1}}{(n+1)!}
\sum_{q=1}^{n+1} \frac{1}{q}.$$
This simplifies to
$$- \frac{1}{n+1} \sum_{q=1}^{n+1} \frac{1}{q}$$
so that
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+k)^2} =
\frac{1}{n+1} \sum_{q=1}^{n+1} \frac{1}{q}
= \frac{1}{n+1} H_{n+1}.}$$
