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$ \sum_{k=1}^\infty k(\frac 14)^k $ i've tried to do the D'Alembert's criterion and i got $ \frac 14 $ but according to wolfram alpha the answer is 4\9

thanks

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We know that $\sum\limits_{i=0}^{\infty}x^i=\frac{1}{1-x} \ \ \text{for all} \ \ |x|<1.$ Therefore $\sum\limits_{i=1}^{\infty}ix^{i-1}=\frac{1}{(1-x)^2} $ and so $\sum\limits_{i=1}^{\infty}ix^{i}=\frac{x}{(1-x)^2} $ now let $x=\frac{1}{4}$ we have $\sum\limits_{i=1}^{\infty}i(\frac{1}{4})^i=\frac{4}{9} $

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You confuse between the ratio test that gives $\frac14<1$ hence we conclude the convergence of the series and the sum of the series $\frac49$ given by the wolfram software.

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