# Question-68554

Question Number 68554 by Mikael last updated on 13/Sep/19
Commented by Prithwish sen last updated on 13/Sep/19
$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{1}}{\mathrm{n}}\left[\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{0}}{\mathrm{n}}}+\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\:+……+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}}\right] \\$$$$=\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\underset{\boldsymbol{\mathrm{r}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}−\mathrm{1}} {\sum}}\:\:\boldsymbol{\mathrm{lim}}\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\:}\frac{\mathrm{1}}{\mathrm{1}+\frac{\boldsymbol{\mathrm{r}}}{\boldsymbol{\mathrm{n}}}}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\boldsymbol{\mathrm{dx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}}\:=\:\boldsymbol{\mathrm{ln}}\left[\mathrm{1}+\boldsymbol{\mathrm{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$
Commented by mathmax by abdo last updated on 13/Sep/19
$${let}\:{A}_{{n}} =\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}\:\Rightarrow{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{n}+{k}} \\$$$$=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}}{{n}}}\:\:{so}\:{A}_{{n}} {is}\:{a}\:{Rieman}\:{sum}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}} \\$$$$=\left[{ln}\mid\mathrm{1}+{x}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:={ln}\left(\mathrm{2}\right)\:. \\$$
Commented by Mikael last updated on 13/Sep/19
$${God}\:{bless}\:{you}\:{Sir}. \\$$
Commented by mathmax by abdo last updated on 13/Sep/19
$${you}\:{are}\:{welcome}. \\$$