Evaluation of $\sum n a^n$ using telescoping property

Question

Show that the series

$$\sum\limits_{n=1}^{\infty}n a^n = \frac{a}{(a-1)^2} $$

for $|a|<1$ using the telescoping property.

I know how to do this using other methods. But the exercise asks to use telescoping property.

Hint: Multiply the series by $(a-1)^2$ and operate the telescoping simplifications. — Siméon, Aug 22 '14 at 11:33
I presume you're meant to proceed as in the first answer here. — David Mitra, Aug 22 '14 at 11:34
let $S= \sum n a^n ,$ then consider $S-aS.$ You can obtain an infinite geometric series. — Bumblebee, Aug 22 '14 at 11:59
@Siméon I got $(na^{n+2} - na^{n+1})-(na^{n+1}-na^n)$, which remember something telescoping but this $n$ is really messing everything — Giiovanna, Aug 22 '14 at 12:10
@Giiovanna: express it only with terms of the form $ka^k$ or $a^k$. For instance $na^{n+1} = (n+1)a^{n+1} - a^{n+1}$. — Siméon, Aug 22 '14 at 12:32
Hi, this series is derivative of the geometric series $\sum_{n=0}^{\infty}a^n=\frac{1}{1-a}$ — M.R. Yegan, Aug 22 '14 at 14:04

score 2 · Accepted Answer · answered Aug 22 '14 at 13:53

From the form of the end result, you want to put $\sum na^n(a-1)$ in the form of a geometric series. This can be done as follows: \begin{align*} \sum_{n\geq 1} na^n(a-1)&=\left(\sum_{n\geq 1}na^{n+1} - (n-1)a^n\right)-\sum_{n\geq 1}a^n\\ &=\left(\lim_{n\rightarrow\infty} na^{n+1} \right)- \sum_{n\geq 1}a^n=\frac{a}{a-1}. \end{align*}

score 0 · Answer 2 · answered Aug 22 '14 at 14:58

$$ \sum_{n\geq 1} (a-1)n a^n = \sum [(n+1)a^{n+1} - na^n] - a^{n+1} $$

We have that

$$\sum a^{n+1} = \frac{a^2}{1-a}$$

and

$$\sum [(n+1)a^{n+1} - na^n] = -a$$

applying the telescoping property. Then,

$$ \sum (a-1) na^n = -a + \frac{a^2}{(a-1)} = \frac{a}{a-1}$$

Then,

$$ \sum na^n= \frac{a}{(a-1)^2}$$

Evaluation of $\sum n a^n$ using telescoping property

2 Answers2