Question Number 136758 by mathlove last updated on 25/Mar/21 | ||
$$\:\:\mathrm{1}==>\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\pi}+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{\mathrm{1}}{\pi^{\mathrm{3}} }+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\pi^{{n}} }\right)=? \\ $$$$ \\ $$$$\:\mathrm{2}=>\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\centerdot\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} }{\mathrm{1}−{n}^{\mathrm{3}} }=? \\ $$ | ||
Answered by Dwaipayan Shikari last updated on 25/Mar/21 | ||
$$\left(\mathrm{1}\right)\:\:\:\frac{\mathrm{1}}{\pi}+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }+...=\frac{\frac{\mathrm{1}}{\pi}}{\mathrm{1}−\frac{\mathrm{1}}{\pi}}=\frac{\mathrm{1}}{\pi−\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +..+{n}^{\mathrm{2}} }{\mathrm{1}−{n}^{\mathrm{3}} }=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}\left(\mathrm{1}−{n}^{\mathrm{3}} \right)}\sim−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}{n}^{\mathrm{3}} } \\ $$$$=−\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{2}+\frac{\mathrm{1}}{{n}}\right)}{\mathrm{6}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$ | ||