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Question Number 82307 by Power last updated on 20/Feb/20

Commented by Tony Lin last updated on 20/Feb/20

$$\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{2}{(n-1)(n+1)}$$$$=\underset{n=3}{\sum ^{\mathrm{\infty }}}(\frac{1}{n-1}-\frac{1}{n+1})$$$$=\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\cdot \cdot \cdot$$$$=\frac{5}{6}$$

Commented by Power last updated on 20/Feb/20

$$\mathrm{thanks}$$

Commented by mathmax by abdo last updated on 20/Feb/20

$$let{S}_n=\sum _{k=0}^n\frac{2}{(k+2)(k+4)}\Rightarrow S_n=\sum _{k=0}^n(\frac{1}{k+2}-\frac{1}{k+4})$$$$=\sum _{k=0}^n\frac{1}{k+2}-\sum _{k=0}^n\frac{1}{k+4}but$$$$\sum _{k=0}^n\frac{1}{k+2}=\frac{1}{2}+\frac{1}{3}+\sum _{k=2}^n\frac{1}{k+2}=\frac{5}{6}+\sum _{p=0}^{n-2}\frac{1}{p+4}(k=p+2)$$$$\Rightarrow S_n=\frac{5}{6}+\sum _{p=0}^{n-2}\frac{1}{p+4}-\sum _{p=0}^n\frac{1}{p+4}$$$$=\frac{5}{6}-\frac{1}{n+3}-\frac{1}{n+4}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{5}{6}$$

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