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Question Number 33124 by prof Abdo imad last updated on 10/Apr/18

$$find\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)2^{n-1}}.$$

Commented by prof Abdo imad last updated on 15/Apr/18

$$letconsiderS\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n(n+1)}with\mid x\mid <1$$$$S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{1}{n+1})x^n$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}but$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}=-ln\mid 1-x\mid and$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}=\sum _{n=2}^{\mathrm{\infty }}\frac{x^{n-1}}{n}=\frac{1}{x}\sum _{n=2}^{\mathrm{\infty }}\frac{x^n}{n}$$$$=\frac{1}{x}(\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-x)=\frac{1}{x}\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-1$$$$=-\frac{ln\mid 1-x\mid }{x}-1\Rightarrow$$$$S\left(x\right)=-ln\mid 1-x\mid +\frac{ln\mid 1-x\mid }{x}+1$$$$=\frac{1-x}{x}ln\mid 1-x\mid -1forx=\frac{1}{2}weget$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)2^n}=2(1-\frac{1}{2})ln\left(\frac{1}{2}\right)+1$$$$=-ln\left(2\right)+1\Rightarrow \frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{(n+1)}^{}{2}^{n-1}}=1-ln\left(2\right)$$$$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)2^{n-1}}=2-2ln\left(2\right)$$

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