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Question Number 67996 by TawaTawa last updated on 03/Sep/19

Commented by mathmax by abdo last updated on 03/Sep/19

$$wehave1⩽k⩽n\Rightarrow n^2+1⩽n^2+k⩽n^2+n\Rightarrow$$$$\sqrt{n^2+1}⩽\sqrt{n^2+k}⩽\sqrt{n^2+n}\Rightarrow \frac{1}{\sqrt{n^2+n}}⩽\frac{1}{\sqrt{n^2+k}}⩽\frac{1}{\sqrt{n^2+1}}\Rightarrow$$$$\frac{n}{\sqrt{n^2+n}}⩽\sum _{k=1}^n\frac{1}{\sqrt{n^2+k}}⩽\frac{n}{\sqrt{n^2+1}}wehave$$$${lim}_{n\to +\mathrm{\infty }}\frac{n}{\sqrt{n^2+n}}={lim}_{n\to +\mathrm{\infty }}\frac{n}{n\sqrt{1+\frac{1}{n}}}={lim}_{n\to +\mathrm{\infty }}\frac{1}{\sqrt{1+\frac{1}{n}}}=1$$$${lim}_{n\to +\mathrm{\infty }}\frac{n}{\sqrt{n^2+1}}={lim}_{n\to +\mathrm{\infty }}\frac{n}{n\sqrt{1+\frac{1}{n^2}}}={lim}_{n\to +\mathrm{\infty }}\frac{1}{\sqrt{1+\frac{1}{n^2}}}=1$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{a}_n=1.$$

Commented by mathmax by abdo last updated on 03/Sep/19

$$b_n=\sum _{k=1}^n\frac{1}{\sqrt{n+k}}wehave1⩽k⩽n\Rightarrow n+1⩽n+k⩽2n\Rightarrow$$$$\sqrt{n+1}⩽\sqrt{n+k}⩽\sqrt{2n}\Rightarrow \frac{1}{\sqrt{2n}}⩽\frac{1}{\sqrt{n+k}}⩽\frac{1}{\sqrt{n+1}}\Rightarrow$$$$\frac{1}{\sqrt{n+k}}⩾\frac{1}{\sqrt{2n}}but{lim}_{n\to +\mathrm{\infty }}\frac{1}{\sqrt{2n}}=+\mathrm{\infty }\Rightarrow {lim}_{n\to +\mathrm{\infty }}{b}_n=+\mathrm{\infty }$$$$c_n=\frac{1}{n}\sum _{k=n}^{2n}\frac{1}{k}changementofindicek-n=igive$$$$c_n=\frac{1}{n}\sum _{i=0}^n\frac{1}{n+i}=\frac{1}{n}\left\{\frac{1}{n}\sum _{i=0}^n\frac{1}{1+\frac{i}{n}}\right\}wehave$$$${lim}_{n\to +\mathrm{\infty }}\frac{1}{n}\sum _{i=0}^n\frac{1}{1+\frac{i}{n}}={\int }_0^1\frac{dx}{1+x}={[ln\mid 1+x\mid ]}_0^1=ln\left(2\right)\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{c}_n=0$$

Commented by TawaTawa last updated on 03/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Abdo msup. last updated on 04/Sep/19

$$youarewelcome.$$

Answered by mind is power last updated on 03/Sep/19

$$a_n=\sum _{k=1}^n\frac{1}{\sqrt{n^2+k}}$$$$\mathrm{\forall }k\in [1,n]$$$$\frac{1}{\sqrt{n^2+n}}⩽\frac{1}{\sqrt{n^2+k}}⩽\frac{1}{\sqrt{n^2+1}}$$$$\Rightarrow \sum _{k=1}^n\frac{1}{\sqrt{n^2+n}}⩽a_n⩽\sum _{k=1}^n\frac{1}{\sqrt{n^2+1}}$$$$\Rightarrow \frac{n}{\sqrt{n^2+n}}⩽a_n⩽\frac{n}{\sqrt{n^2+1}}$$$$\Rightarrow {lima}_n=1$$$$\sum _{k=1}^n\frac{1}{\sqrt{n+k}}=b_n$$$$\mathrm{\forall }k\frac{1}{\sqrt{n+k}}⩾\frac{1}{\sqrt{n+n}}=\frac{1}{\sqrt{2n}}$$$$\Rightarrow \sum _{k=1}^n\frac{1}{\sqrt{n+k}}⩾\mathrm{\Sigma }\frac{1}{\sqrt{2n}}=\frac{n}{\sqrt{2n}}=\sqrt{\frac{n}{2}}$$$$\Rightarrow b_n\to +\mathrm{\infty }$$$$c_n=\frac{1}{n}\sum _n^{2n}\frac{1}{k}=\frac{1}{n_{}}\sum _{k=n}^{k=2n}\frac{1}{k}$$$$weshow\sum _{k=n}^{k=2n}\frac{1}{k_{}}cv\Rightarrow c_n\to 0$$$$\sum _n^{2n}\frac{1}{k}=\sum _{k=0}^n\frac{1}{2n-k}=\frac{1}{n}\sum _{k=0}^n\frac{1}{2-\frac{k}{n}}$$$$letf\left(x\right)=\frac{1}{2-x}$$$$\Rightarrow \frac{1}{n}\sum _{k=0}^n\frac{1}{2-\frac{k}{n}}=\frac{1}{n}\sum _{k=0}^{n}f\left(\frac{k}{n}\right)={\int }_0^{1}f\left(x\right)dxreiman$$$${\int }_0^{1}f\left(x\right)dx={\int }_0^1\frac{1}{2-x}dx={[-ln(2-x)]}_0^1=ln\left(2\right)$$$$\Rightarrow \sum _{k=n}^{2n}\frac{1}{k}\to ln\left(2\right)$$$$\Rightarrow \frac{1}{n}\sum _{k=n}^{k=2n}\to 0$$$$$$$$$$

Commented by TawaTawa last updated on 03/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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