Question Number 79625 by M±th+et£s last updated on 26/Jan/20 | ||
$${if}\:{n}>\mathrm{1}\:{prove}\:{that} \\ $$ $$\mathrm{2}{ln}\left({n}\right)−{ln}\left({n}+\mathrm{1}\right)−{ln}\left({n}−\mathrm{1}\right)=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{6}} }+...= \\ $$ | ||
Answered by mind is power last updated on 27/Jan/20 | ||
$$\Leftrightarrow{ln}\left(\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} −\mathrm{1}}\right)={ln}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}\right)=−{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$ $$\forall{n}\in\mathbb{N}−\left\{\mathrm{0},\mathrm{1}\right\}\:\:\:\frac{\mathrm{1}}{{n}}<\mathrm{1}\Rightarrow\frac{\mathrm{1}}{{n}^{\mathrm{2}} }<\mathrm{1} \\ $$ $$\left.−{ln}\left(\mathrm{1}−{x}\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{k}} }{{k}},\forall{x}\in\right]−\mathrm{1},\mathrm{1}\left[,{x}=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\Rightarrow\right. \\ $$ $$−{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}{k}} .{k}}=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }+.......... \\ $$ $$ \\ $$ | ||