Question Number 30758 by abdo imad last updated on 25/Feb/18
$${find}\:{lim}_{{n}\rightarrow\infty} \left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:+….+\frac{\mathrm{1}}{{n}+{p}}\right)\:{pfixed}\:{fromN}^{\bigstar} \\ $$$$ \\ $$
Commented by abdo imad last updated on 28/Feb/18
$${let}\:{put}\:{u}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:+…+\frac{\mathrm{1}}{{n}+{p}} \\ $$$${u}_{{n}} ={H}_{{n}+{p}} \:−{H}_{{n}} ={ln}\left({n}+{p}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:−{ln}\left({n}\right)\:−\gamma\:−{o}^{'} \left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$={ln}\left(\frac{{n}+{p}}{{n}}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)_{{n}\rightarrow\infty} \rightarrow\mathrm{0}\:\Rightarrow\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} =\mathrm{0} \\ $$