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Question Number 139532 by mathdanisur last updated on 28/Apr/21

$$\frac{1}{3}-\frac{x-3}{9}+\frac{{(x-3)}^2}{27}+...=?$$

Commented by mr W last updated on 28/Apr/21

$$=\frac{1}{x}$$

Commented by mathdanisur last updated on 28/Apr/21

$$howsir?$$

Commented by mr W last updated on 28/Apr/21

$$a_0+a_{0}q+a_0{q}^2+...=\frac{a_0}{1-q}with\mid q\mid <1$$$$now{a}_0=\frac{1}{3},q=-\frac{x-3}{3}$$$$\frac{1}{3}-\frac{x-3}{9}+\frac{{(x-3)}^2}{27}+...$$$$=\frac{1}{3}\times \frac{1}{1-(-\frac{x-3}{3})}=\frac{1}{x}$$

Commented by mathdanisur last updated on 30/Apr/21

$$thankssir$$

Answered by mathmax by abdo last updated on 28/Apr/21

$$\mathrm{S}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{(\mathrm{x}-3)}^{\mathrm{n}}}{3^{\mathrm{n}+1}}=\mathrm{w}(\mathrm{x}-3)\mathrm{withw}\left(\mathrm{u}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}}{3^{\mathrm{n}+1}}$$$$\mathrm{we}\mathrm{have}\mathrm{w}\left(\mathrm{u}\right)=\frac{1}{3}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\left(\frac{\mathrm{u}}{3}\right)}^{\mathrm{n}}=\frac{1}{3}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-\frac{\mathrm{u}}{3})}^{\mathrm{n}}$$$$=\frac{1}{3}\frac{1}{1+\frac{\mathrm{u}}{3}}=\frac{1}{\mathrm{u}+3}\Rightarrow \mathrm{S}\left(\mathrm{x}\right)=\mathrm{w}(\mathrm{x}-3)=\frac{1}{\mathrm{x}-3+3}=\frac{1}{\mathrm{x}}\Rightarrow$$$$\mathrm{S}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\mathrm{and}\mathrm{condition}\mathrm{of}\mathrm{convergence}\mid \mathrm{u}\mid <3\Rightarrow \mid \mathrm{x}-3\mid <3$$

Commented by mathdanisur last updated on 30/Apr/21

$$thankssir$$

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