Question Number 170003 by mathlove last updated on 13/May/22
$${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Commented by mathlove last updated on 14/May/22
$${any}\:{one}\:{solve}\:{this} \\ $$
Answered by ajfour last updated on 14/May/22
$$\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}=\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{x}^{\mathrm{3}} }{\mathrm{4}}+.. \\ $$$$\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\:^{\mathrm{1}} } \:\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right){dx}}{{x}}=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+.. \\ $$$$={I}=\int_{\mathrm{0}} ^{\:^{\mathrm{1}} } \:\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right){dx}}{{x}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\: \\ $$