IRL_ABooklet

Let $H_n={\sum }_{k=1}^n\frac{1}{k}$, then $\frac{H_{m+1}-H_m}{m}=\frac{1}{m(m+1)}=\frac{1}{m}-\frac{1}{m+1}$, and $$\begin{array}{rl}\sum _{m=1}^M\frac{1}{m(m+1)}\sum _{k=1}^{m+1}\frac{1}{k}& =\sum _{m=1}^M\frac{1}{m(m+1)}{H}_{m+1}\\ & =\sum _{m=1}^M\left(\frac{1}{m}-\frac{1}{m+1}\right)H_{m+1}\\ & =\sum _{m=1}^M\frac{1}{m}{H}_{m+1}-\sum _{m=1}^M\frac{1}{m+1}{H}_{m+1}\\ & =H_2+\sum _{m=2}^M\frac{1}{m}{H}_{m+1}-\sum _{m=2}^{M+1}\frac{1}{m}{H}_m\\ & =\frac{3}{2}+\sum _{m=2}^M\frac{1}{m}(H_{m+1}-H_m)-\frac{1}{M+1}{H}_{M+1}\\ & <\frac{3}{2}+\sum _{m=2}^M\left(\frac{1}{m}-\frac{1}{m+1}\right)=2-\frac{1}{M+1}<2.\end{array}$$