Let $m$ be a positive integer, then $S=\sum _{k=1}^m k\left(\frac{1}{k}+\frac{1}{k+1}+\frac{1}{k+2}+\dots +\frac{1}{m}\right)$ is equal to :

$\begin{array}{l}S=\left(1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{m}\right)\\ +2\left(\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{m}\right)\\ +3\left(\frac{1}{3}+\frac{1}{4}+\dots +\frac{1}{m}\right)\\ +\dots \\ +(m-1)\left(\frac{1}{m-1}+\frac{1}{m}\right)+m\left(\frac{1}{m}\right)\\ =\left(1\right)\left(1\right)+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)\\ +\frac{1}{4}(1+2+3+4)\\ +\dots +\frac{1}{m}(1+2+3+\dots +m)\end{array}$

$\begin{array}{l}=\sum _{k=1}^m \frac{1}{k}(1+2+\dots +k)\\ =\sum _{k=1}^m \frac{1}{k}\cdot \frac{k(k+1)}{2}=\frac{1}{2}\sum _{k=1}^m (k+1)\\ =\frac{m}{4}[2+(m+1\left)\right]\\ =\frac{1}{4}m(m+3)\end{array}$