How to prove $\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \frac{1}{N+1} \sum\limits_{n=1}^{N+1} \frac{1}{n}$

Question

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.

Answer 1

\begin{align} (N+1)\sum_{k=0}^N\frac{(-1)^k}{(k+1)^2}\binom Nk &=\sum_{k=0}^N\frac{(-1)^k}{k+1}\binom{N+1}{k+1}\\ &=-\sum_{k=0}^N(-1)^{k+1}\binom{N+1}{k+1}\int_0^1t^k\,dt\\ &=\int_0^1\frac{1-(1-t)^{N+1}}{t}\,dt\\ &=\int_0^1\frac{1-u^{N+1}}{1-u}\,du\\ &=\int_0^1\sum_{n=1}^{N+1}u^{n-1}\,du\\ &\sum_{n=1}^{N+1}\frac1n \end{align} where I substituted $u=1-t$.

Answer 2

Since $\frac{1}{(k+1)^2}=\int_{0}^{1}x^k(-\log x)\,dx $ we have $$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^2}=\int_{0}^{1}(-\log x)\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx=\int_{0}^{1}(-\log x)(1-x)^n\,dx. $$ If we perform the substitution $x\mapsto 1-x$ in the last integral we get $$ \int_{0}^{1} x^n(-\log(1-x))\,dx=\int_{0}^{1}\sum_{k\geq 1}\frac{x^{n+k}}{k}\,dx=\sum_{k\geq 1}\frac{1}{k(n+k+1)}=\frac{1}{n+1}\sum_{k\geq 1}\left(\frac{1}{k}-\frac{1}{n+k+1}\right) $$ and by telescopic terms the RHS equals $\frac{H_{n+1}}{n+1}$ QED.

Answer 3

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

$$\bbox[5px,border:2px solid #00A000]{ \sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+k)^2} = \frac{1}{n+1} \sum_{q=1}^{n+1} \frac{1}{q} = \frac{1}{n+1} H_{n+1}.}$$

Answer 4

It is a special case of the derivative version of the Melzak's identity. Let $f$ be an algebraic polynomial up to degree $N$ and let $x\neq-k,\,k\in\mathbb{N}$. Then $$ \sum_{k=0}^{N}\left(-1\right)^{k}\dbinom{N}{k}\frac{f\left(y-k\right)}{\left(x+k\right)^{2}}=\frac{f\left(x+y\right)\sum_{k=0}^{N}\frac{1}{x+k}-\frac{d}{dx}f\left(x+y\right)}{x\dbinom{x+N}{N}}.$$ Now take $f\equiv1$ and $x=1$.