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Question Number 68554 by Mikael last updated on 13/Sep/19

Commented by Prithwish sen last updated on 13/Sep/19

$$\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{1}{\mathrm{n}}[\frac{1}{1+\frac{0}{\mathrm{n}}}+\frac{1}{1+\frac{1}{\mathrm{n}}}+......+\frac{1}{1+\frac{\mathrm{n}-1}{\mathrm{n}}}]$$$$=\frac{1}{\mathbf{n}}\underset{\mathbf{r}=0}{\sum ^{\mathbf{n}-1}}\mathbf{lim}\underset{\mathbf{n}\to \mathrm{\infty }}{}\frac{1}{1+\frac{\mathbf{r}}{\mathbf{n}}}=\underset{0}{\stackrel{1}{\int }}\frac{\mathbf{dx}}{1+\mathbf{x}}=\mathbf{ln}{[1+\mathbf{x}]}_0^1$$

Commented by mathmax by abdo last updated on 13/Sep/19

$$let{A}_n=\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{3n-1}\Rightarrow A_n=\sum _{k=0}^{n-1}\frac{1}{n+k}$$$$=\frac{1}{n}\sum _{k=0}^{n-1}\frac{1}{1+\frac{k}{n}}so{A}_{n}isaRiemansum\Rightarrow {lim}_{n\to +\mathrm{\infty }}{A}_n={\int }_0^1\frac{dx}{1+x}$$$$={[ln\mid 1+x\mid ]}_0^1=ln\left(2\right).$$

Commented by Mikael last updated on 13/Sep/19

$$GodblessyouSir.$$

Commented by mathmax by abdo last updated on 13/Sep/19

$$youarewelcome.$$

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