# Find $\sum \limits_{k=1}^{\infty} k q^{k-1} \,\,$ when $\,\, 0 < q < 1$ [duplicate]

Possible Duplicate:
Sequence sum question

Please find $$\sum \limits_{k=1}^{\infty} k q^{k-1}$$ when $$0 < q < 1$$.

This was an exam question yesterday. I thought on it a lot during the exam, but couldn't solve it. And it is frustrating that I haven't still solved it.

I found this similar identity in a Wikipedia page:

$$\sum_{i=0}^{n-1} i x^i = \frac{x-nx^n+(n-1)x^{n+1}}{(1-x)^2}$$

But I am not able to apply it to my problem.

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## marked as duplicate by lhf, Michael Albanese, Per Manne, rschwieb, Ittay WeissJan 31 '13 at 11:44

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.

## 2 Answers

This is called Arithmetico-geometric series or sequence.

Let $$S=\sum_{1\le k \le n}kq^{k-1}=1+2q+3q^2+\cdots+(n-1)q^{n-2}+(n)q^{n-1}$$ where $q\ne1$

So, $$qS=q\sum_{1\le k \le n}kq^{k-1}=q+2q^2+3q^3+\cdots+(n-1)q^{n-1}+(n)q^{n}$$

On subtraction, $$(1-q)S=1+(2q-q)+(3q^2-2q^2)+\cdots+\{n-(n-1)\}q^{n-1}-nq^n$$ $$=-nq^n+1+q+q^2+\cdots+q^{n-1}=-nq^n+\frac{1-q^n}{1-q}$$

So, $$S=\frac{1-q^n}{(1-q)^2}-\frac{nq^n}{1-q}$$

If $0<q<1$ and $n\to\infty,q^n=0$ and from here $\lim_{n\to\infty}nq^n=0$

So, $$\lim_{n\to\infty}S=\frac1{(1-q)^2}$$

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$$f(x):=\sum_{i=0}^\infty x^i=\frac{1}{1-x}\,\,,\,\,|x|<1\Longrightarrow\frac{1}{(1-x)^2}=f'(x)=\sum_{k=1}^\infty ix^{i-1}\,\,,\,\,|x|<1\ldots$$

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