Integration Questions

Question Number 101791 by Dwaipayan Shikari last updated on 04/Jul/20

$$\frac{\mathrm{1}}{{n}^{\mathrm{3}\:\:} }\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({ne}^{−\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} } +\mathrm{2}{ne}^{−\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} } +....\infty\right) \\$$

Answered by Ar Brandon last updated on 04/Jul/20

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\left\{\mathrm{ne}^{−\left(\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{2}} } +\mathrm{2ne}^{−\left(\frac{\mathrm{2}}{\mathrm{n}}\right)^{\mathrm{2}} } +\centerdot\centerdot\centerdot+\mathrm{n}\centerdot\mathrm{ne}^{−\left(\frac{\mathrm{n}}{\mathrm{n}}\right)^{\mathrm{2}} } \right\} \\$$$$\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left\{\mathrm{kne}_{} ^{−\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} } \right\}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left\{\frac{\mathrm{k}}{\mathrm{n}}\mathrm{e}^{−\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} } \right\} \\$$$$\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\frac{\mathrm{1}}{−\mathrm{2}}\left[\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}}\right\} \\$$$$\\$$

Commented by Dwaipayan Shikari last updated on 04/Jul/20

$${You}\:{are}\:{right}\:{sir}\:.\:{It}\:{will}\:{be}\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\$$

Answered by mathmax by abdo last updated on 04/Jul/20

$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{nk}\:\mathrm{e}^{−\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} \:}} \:\Rightarrow\:\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{n}}\mathrm{e}^{−\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} } \:\Rightarrow \\$$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}}\right) \\$$$$=\frac{\mathrm{e}−\mathrm{1}}{\mathrm{2e}} \\$$

Commented by Dwaipayan Shikari last updated on 04/Jul/20

$${Thanking}\:{you} \\$$

Commented by mathmax by abdo last updated on 05/Jul/20

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\$$