0
$\begingroup$

This question already has an answer here:

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

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

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4
$\begingroup$

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 $$ \sum_{n=0}^\infty n \left(\frac23\right)^{n}=\frac{\frac23}{\left(1-\frac23\right)^2}=\large\color{blue}{6}. $$

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.