Evaluating $ \sum_{i=0}^{\infty}ir^i$ for $|r|\lt 1$

Question

I'm doing my mathematics homework and there's question which I'm pretty much unable to solve. I've literally tried every method but no results. It'd be great, if anybody could help me. Thanks (in advance).

$$ \sum_{i=0}^{\infty}ir^i$$ $$|r|\lt 1$$I need the step-wise procedure.

I tried solving it on wolfram alpha and the answer which it gives it, $$\frac{r}{(1-r)^2}$$

What does "mod(r)<1" mean? Perhaps you meant absolute value, i.e. $;|r|<1;$ ? — Timbuc, Oct 19 '14 at 12:27

score 2 · Answer 1 · answered Oct 19 '14 at 12:29

2

Hint (almost no complete, stepwise solutions here):

$$|x|<1\implies \frac1{1-x}=\sum_{k=0}^\infty x^k\implies\;\text{for}\;\;|x|<1\,,\;\; \left(\frac1{1-x}\right)'=\sum_{k=1}^\infty kx^{k-1}$$

answered Oct 19 '14 at 12:29

Timbuc

34,191

Evaluating $ \sum_{i=0}^{\infty}ir^i$ for $|r|\lt 1$

1 Answers1