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Question Number 29986 by abdo imad last updated on 14/Feb/18

$$find\sum _{n=0}^{\mathrm{\infty }}\frac{n+1}{4^n}.$$

Commented by abdo imad last updated on 14/Feb/18

$$letintroducefor\mid x\mid <1S\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}(n+1)x^{n}wehave$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{n+1}{4^n}=S\left(\frac{1}{4}\right)wehave{\int }_0^{x}S\left(t\right)dt=\sum _{n=0}^{\mathrm{\infty }}{x}^{n+1}+\lambda$$$$x=0\Rightarrow \lambda =0and{\int }_0^{x}S\left(t\right)dt=\sum _{n=1}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}-1=\frac{x}{1-x}$$$$\Rightarrow S\left(x\right)=\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{1-x-x(-1)}{{(1-x)}^2}=\frac{1}{{(1-x)}^2}\Rightarrow$$$$S\left(\frac{1}{4}\right)=\frac{1}{{\left(\frac{3}{4}\right)}^2}=\frac{16}{9}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{n+1}{4^n}=\frac{16}{9}.$$

Answered by MJS last updated on 14/Feb/18

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n+1}{k^n}=\frac{k^2}{{(k-1)}^2}\mathrm{with}k>1$$

Commented by abdo imad last updated on 14/Feb/18

$$youmustshowyourworkandyourmethodsirmjs...$$

Commented by abdo imad last updated on 15/Feb/18

$$wehaveprovedthat\sum _{n=0}^{\mathrm{\infty }}(n+1)x^n=\frac{1}{{(1-x)}^2}for\mid x\mid <1$$$$andforx=\frac{1}{k}withk>1weobtain$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{n+1}{k^n}=\frac{1}{{(1-\frac{1}{k})}^2}=\frac{k^2}{{(k-1)}^2}.$$

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