How to calculate complicated geometrical series?

Question

I have a geometrical series (I don't know if its geometrical series or not):

$$ \sum_{n=1}^{\infty }n\rho ^{n}(1-\rho) $$

how can I simplify it ? ( assume that $ 0 \le \rho \le 1$ )

The last answer in my calculatio should be $\frac{\rho}{1-\rho}$. But I really don't know how ?

It's not Geometric. It is almost a "Differentiated Geometric series", though. See this for many ways to calculate the sum of such a series. — David Mitra, Jan 10 '14 at 14:52

score 4 · Accepted Answer · answered Jan 10 '14 at 14:51

4

$$\sum_{n=1}^{\infty }n\rho ^{n}(1-\rho)=(1-\rho)\rho\frac{d(\sum_{n=1}^{\infty }\rho^n)}{d \rho}$$

Now using Infinite Geometric Series $$\sum_{n=1}^{\infty }\rho^n=\frac \rho{1-\rho}=-1-\frac1{\rho-1}$$ as $|\rho|<1$

answered Jan 10 '14 at 14:51

lab bhattacharjee

How did you do that ? I mean how first statement created ?? I don't get it – Michel Gokan Khan Jan 10 '14 at 14:55
@linker, do you know $$\frac{d(x^n)}{dx}=?$$ – lab bhattacharjee Jan 10 '14 at 14:56
$nx ^{n-1}$ ? so ? whats diffrential ? where did it came from ? – Michel Gokan Khan Jan 10 '14 at 14:57
@linker, $$\sum_{n=1}^{\infty }n\rho ^n(1-\rho)=(1-\rho)\rho\sum_{n=1}^{\infty }n\rho ^{n-1}= $$ – lab bhattacharjee Jan 10 '14 at 14:59
You have made a small mistake. The Geometric is : $\sum_{n=0}^{\infty }\rho^n=\frac{1}{1-\rho}$ – Jan 10 '14 at 15:08
@Julien, please check http://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series, here $a=\rho$ – lab bhattacharjee Jan 10 '14 at 15:09
@Julien no he did it right, you are right to when n starts from 0 – Michel Gokan Khan Jan 10 '14 at 15:09
@labbhattacharjee you are awesome :) – Michel Gokan Khan Jan 10 '14 at 15:10
Yes, sorry I read too fast.. – Jan 10 '14 at 15:10
@linker, thanks for your feedback. Btw, have you looked into my other solution? – lab bhattacharjee Jan 10 '14 at 15:11
@labbhattacharjee I really don't know that kind of series, I will take a look it later thnaks anyway <3 – Michel Gokan Khan Jan 10 '14 at 15:30

lab bhattacharjee · Answer 2 · 2014-01-10T15:32:59.910

1

Using Arithmetico-geometric sequence, $$\sum_{n=1}^\infty\left[a+(n-1)d\right]r^{n-1}=\frac a{1-r}+\frac{rd}{(1-r)^2}\text{ for } |r|<1$$

We have $$\sum_{n=1}^{\infty}n\rho^n(1-\rho)=(1-\rho)\rho\left(\sum_{n=1}^{\infty}n \rho^{n-1}\right)$$

Here, we have $\displaystyle a=d=1,r=\rho$

edited Jan 10 '14 at 15:32

answered Jan 10 '14 at 15:03

lab bhattacharjee

@Dwonvoter, please have a look into the edited version. Wish I could know the mistake here – lab bhattacharjee Jan 10 '14 at 15:35

2 Answers2