Question Number 28891 by abdo imad last updated on 31/Jan/18
$${let}\:{give}\:{u}_{{n},{k}} =\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:+....\:\frac{\mathrm{1}}{{kn}}\:\:\:{k}\:{integr}\:{fixed}\:\geqslant\mathrm{2} \\ $$$${find}\:{lim}_{{n}\rightarrow+\:\:\infty} {u}_{{n},{k}} . \\ $$
Answered by ajfour last updated on 01/Feb/18
$${u}=\underset{{r}=\mathrm{1}} {\overset{{n}\left({k}−\mathrm{1}\right)} {\sum}}\frac{\mathrm{1}/{n}}{\mathrm{1}+\frac{{r}}{{n}}}\:=\:\int_{\mathrm{0}} ^{\:\:{k}−\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}} \\ $$$$=\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\mid_{\mathrm{0}} ^{{k}−\mathrm{1}} \:=\:\mathrm{ln}\:\boldsymbol{{k}}\:. \\ $$