Find the sum of infinite terms of the series | vedclass

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(D) Let the sum be $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \cdots$This is an arithmetico-geometric series of the form $

Vedclass · Accepted Answer

$\frac{9}{4}$

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(D) Let the sum be $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \cdots$This is an arithmetico-geometric series of the form $S = \sum_{n=1}^{\infty} n x^{n-1}$ where $x = \frac{1}{3}$.The sum of this series is given by the formula $S = \frac{1}{(1-x)^2}$ for $|x| Substituting $x = \frac{1}{3}$ into the formula:$S = \frac{1}{(1 - \frac{1}{3})^2}$$S = \frac{1}{(\frac{2}{3})^2}$$S = \frac{1}{\frac{4}{9}}$$S = \frac{9}{4}$

Find the sum of infinite terms of the series $1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\frac{5}{3^{4}}+\cdots$

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