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$$\sum_{n=0}^{\infty} n \left( \frac{2}{3} \right)^n = ?$$

How to find it? If it lacked n before fraction, I would use formula for the sum of geometric series.


marked as duplicate by user61527, lab bhattacharjee calculus Jul 5 '14 at 10:04

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Consider the infinite geometric progression for $|x|<1$, $$ \sum_{n=0}^\infty x^n=\frac1{1-x}.\tag1 $$ Differentiating $(1)$ with respect to $x$ yields $$ \sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}.\tag2 $$ Multiplying $(2)$ by $x$ yields $$ \sum_{n=0}^\infty nx^{n}=\frac{x}{(1-x)^2}.\tag3 $$ Setting $x=\dfrac23$ to $(3)$ yields $$ \sum_{n=0}^\infty n \left(\frac23\right)^{n}=\frac{\frac23}{\left(1-\frac23\right)^2}=\large\color{blue}{6}. $$

  • $\begingroup$ (1) is true for $\;|x|<1\;$ $\endgroup$ – Timbuc Jul 5 '14 at 10:03

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