\sum\displaylimits_{k=1}^{m+1}\frac{1}{\sqrt{k}}>\sqrt{m}+\frac{1}{\sqrt{m+1}}=\sqrt{m+1}\Biggl(\sqrt{\frac{m}{m+1}}+\frac{1}{m+1}\Biggr)=\sqrt{m+1}\Biggl[\frac{\sqrt{m(m+1)}}{m+1}+\frac{1}{m+1}\Biggr]=