I have the series 1r, 2r^2,3r^3...(or in general nr^n)
what is the sum of this series? Im a bit confused as it is a mixture of an arithmetic series (1,2,3...) and a geometric (r^n)
Thanks for your help
Is...
$\displaystyle \sum_{n=1}^{\infty} n\ r^{n} = r\ \sum_{n=1}^{\infty} n\ r^{n-1} = r\ \frac{d}{d r}\ \sum_{n=1}^{\infty} r^{n} $ (1)
... and, because for $\displaystyle |r|<1$...
$\displaystyle \sum_{n=1}^{\infty} r^{n} = \frac{1}{1-r} - 1 $ (2)
... we have for $\displaystyle |r|<1$ ...
$\displaystyle \sum_{n=1}^{\infty} n\ r^{n} = \frac{r}{(1-r)^{2}}$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
You can get to a standard geometric series as follows....
$\displaystyle S_n=r+2r^2+3r^3+.....+nr^n$
$\displaystyle rS_n=r^2+2r^3+3r^4+....+nr^{n+1}$
$\displaystyle S_n-rS_n=r+r^2+r^3+....r^n-nr^{n+1}$
$\displaystyle S_n(1-r)=\left(r+r^2+r^3+...+r^n\right)-nr^{n+1}$
The geometric series in brackets can now be evaluated
and hence a closed form for the sum can be found.
Another way: denoting $\displaystyle S=r+2r^2+3r^3+\ldots\quad$ , we have $\displaystyle rS=r^2+2r^3+3r^3+\ldots$ . This implies
$\displaystyle (1-r)S=r+r^2+r^3+\ldots=\dfrac{1}{1-r}-1=\dfrac{r}{1-r}\quad (|r|<1)$
So, $\displaystyle S=\dfrac{r}{(1-r)^2}$
Edited: Sorry, I didn't see Archie Meade's post