Binomial sum

Let $\mathbb{N} \ni k >1$ . Evaluate the sum

$\sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{\binom{n+k}{k}}$

Solution

We have successively

$\begin{align*} \sum_{n=1}^{\infty} \frac{H_n}{\binom{n+k}{k}} &= \sum_{n=1}^{\infty} \frac{k!H_n}{(n+1) \cdots (n+k)} \\ &= \sum_{n=1}^{\infty} \frac{k!H_n}{k-1} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1}\sum_{n=1}^{\infty} \sum_{i=1}^n \frac{1}{i} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1} \sum_{i=1}^{\infty} \frac{1}{i} \sum_{n=i}^{\infty} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1} \sum_{i=1}^{\infty} \frac{1}{i(i+1) \cdots (i+k-1)} \\ &= \frac{k!}{(k-1)^2} \sum_{i=1}^{\infty} \left(\frac{1}{i(i+1) \cdots (i+k-2)} - \frac{1}{(i+1) \cdots (i+k-1)}\right) \\ &= \frac{k!}{(k-1)^2} \frac{1}{(k-1)!} \\ &= \frac{k}{(k-1)^2} \end{align*}$

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