Question Number 82225 by M±th+et£s last updated on 19/Feb/20
Commented by mathmax by abdo last updated on 19/Feb/20
$${let}\:{S}_{{n}} =\sum_{{p}={n}+\mathrm{1}} ^{{kn}} \:\frac{\mathrm{1}}{{p}}\:\Rightarrow{S}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{kn}} \:\frac{\mathrm{1}}{{p}}−\sum_{{p}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{p}}\:={H}_{{kn}} −{H}_{{n}} \\ $$$${we}\:{have}\:{H}_{{kn}} ={ln}\left({kn}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:{also}\:{H}_{{n}} ={ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\left({n}\rightarrow+\infty\right) \\ $$$${H}_{{kn}} −{H}_{{n}} ={ln}\left({k}\right)+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} ={ln}\left({k}\right) \\ $$
Commented by M±th+et£s last updated on 19/Feb/20
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 19/Feb/20
$${you}\:{are}\:{welcome}. \\ $$