TITLE: Harmonic number identity
QUESTION [10 upvotes]: I search for an elementary proof of the following identity:
$$
\sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i}\binom{n}{i+k}=\binom{n}{k}\left(H_n-H_k\right)
$$
I have found the following proof:
$$
\sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i}\binom{n}{i+k}=\binom{n}{k}\left(H_n-H_k\right)\iff\sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i}\frac{k!(n-k)!}{(i+k)!(n-i-k)!}=H_n-H_k
$$
Now:
$$
\sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i}\frac{k!(n-k)!}{(i+k)!(n-i-k)!}=\sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i}\frac{i!\cdot k!}{(i+k)!}\frac{(n-k)!}{i!\cdot(n-i-k)!}=\sum_{i=1}^{n-k} (-1)^{i+1}\frac{\Gamma(i)\cdot \Gamma(k+1)}{\Gamma(i+k+1)}\binom{n-k}{i}={\sum_{i=1}^{n-k} (-1)^{i+1}\cdot\int_{0}^{1} x^{i-1}(1-x)^k dx\cdot\binom{n-k}{i}}={\int_{0}^{1} -\frac{(1-x)^k}{x}\cdot\sum_{i=1}^{n-k} \binom{n-k}{i}(-x)^{i}} \space dx={\int_{0}^{1} -\frac{(1-x)^k}{x}\cdot ((1-x)^{n-k}-1) \space dx}={\int_{0}^{1} \frac{1}{x}\cdot \left((1-x)^{k}-(1-x)^n\right) \space dx}={\int_{0}^{1} \frac{1}{1-x}\cdot \left(x^{k}-x^n\right) \space dx}={\int_{0}^{1} \frac{1-x^n}{1-x}-\frac{1-x^k}{1-x}\space dx}={\int_{0}^{1} \frac{1-x^n}{1-x}\space dx-\int_{0}^{1}\frac{1-x^k}{1-x}\space dx}={\int_{0}^{1} \sum_{i=0}^{n-1}x^i\space dx-\int_{0}^{1}\sum_{i=0}^{k-1}x^i\space dx}={\sum_{i=0}^{n-1}\int_{0}^{1}x^i\space dx-\sum_{i=0}^{k-1}\int_{0}^{1} x^i\space dx}={\sum_{i=0}^{n-1}\frac{1}{i+1}-\sum_{i=0}^{k-1}\frac{1}{i+1}}=H_n-H_k
$$
But think it isn't very elegant since it uses rather advanced techniques like the integral representation of the Beta-function. So is there a more elegant way?

REPLY [6 votes]: I upvoted the excellent answer by @FelixMarin.
Note that we do not need the Gamma function step. We have $$\lim_{\mu\to 0} \frac{\partial}{\partial\mu} {\mu+n\choose n-k}
= \lim_{\mu\to 0} \frac{\partial}{\partial\mu} \frac{1}{(n-k)!} 
\prod_{q=0}^{n-k-1} (\mu+n-q)
\\= \frac{1}{(n-k)!} 
\lim_{\mu\to 0} \prod_{q=0}^{n-k-1} (\mu+n-q) \sum_{q=0}^{n-k-1} \frac{1}{\mu+n-q}
\\= \frac{1}{(n-k)!} 
\frac{n!}{k!} (H_n - H_k) = {n\choose k} (H_n - H_k).$$