Simplify a sum of harmonic numbers

Question

I am dealing with a sum of harmonic numbers: $$ \frac{1}{n-j+1}\sum_{\ell=n-j+2}^{n}\frac{1}{\ell}, \;\;\;2\le j\le n. $$ I wonder if it is possible to simplify it further. Or, is it related to a Riemann sum? (when $n$ is large). Please give me some hints or references. Thanks a lot.

Answer 1

This solution is about the limit when $n\mapsto \infty $:

We have $$\sum_{l=n-j+2}^n\frac1l=\sum_{l=1}^{j-1}\frac1{l+n-j+1}=H_{2n-j+1}-H_{n-j+1}$$

Then

$$\lim_{n\mapsto \infty}\frac{1}{n-j+1}\sum_{l=n-j+2}^n\frac1l=\lim_{n\mapsto \infty}\frac{H_{2n-j+1}-H_{n-j+1}}{n-j+1}$$

Clearly, the denominator blows to infinity much faster than the numerator, so the limit is zero.