The Harmonic number $H_n$ is defined as $H_n=\sum_{t=1}^n\frac{1}{t}$.
I wish to compute $\lim_{n\to\infty}\frac{k}{n}(H_n-H_k)$ where $k$ is a function of $n$ (which can be a constant function, e.g. $k=3$).
Moreover, I wish the computation to be as simple as possible and self-sufficient (which might already be too much to ask). If a general solution is not possible, my main interest is in the case $k=p\cdot n$, with $0<p<1$.
When $n\to+\infty$, $H_n=\log(n)+\gamma+o(1)$ where $\gamma$ is Euler-Mascheroni constant. Using the notation $x(k,n)=(H_n-H_k)k/n$, one deduces the following:
(1) If $k\ll n/\log(n)$ (for example if $k$ is fixed) and $n\to+\infty$, then $x(k,n)\le kH_n/n\to0$ hence $x(k,n)\to0$.
(2) If $k\to+\infty$ and $n\to+\infty$, then $x(k,n)\sim u(n/k)$ with $u(x)=\log(x)/x$ for every $x\ge1$.
(3) If $k\to+\infty$ and $n\to+\infty$ and furthermore $k/n\to p$, then $x(k,n)\to-p\log(p)$.
The answer naturally depends on function $k$. For $k(p)=np$ it is $-p\log p$ since $H_n=\log n+C+O(1/n)$, $n\to\infty\,$, where $C=0.577\ldots$ is the Euler constant.
