Question

I'd like to prove this infinite series formula. Any hints will be very appreciated, thanks.
(Writing the sum in the first post)

Answer 1

\[\sum_{n=1}^{\infty}nr^n=\frac{ r }{ (1-r)^2 }\]

Answer 2

Are you sure it's r and not nr on the numerator?

Answer 3

ok, sorry, it is r, after all. It's all about sorting them, I guess?

Answer 4

You know the formula for the sum of an infinite geometric series?

Answer 5

Ok, so our series takes the form
\[r^{1} + 2r^{2} + 3r^{3} + 4r^{4}...\]
this can be written as
\[r^{1} + r^{2} + r^{3} + r^{4}...\]\[+\]\[r^{2}+r^{3}+r^{4} + r^{5}...\]\[+\]\[r^{3}+r^{4}+r^{5} + r^{6}...\]\[+\]\[r^{4}+r^{5}+r^{6} + r^{7}...\]\[...\]

Each of these is just a geometric series with common ratio r, they only differ in their first terms

Answer 6

Them being geometric series, their sums are given by...
\[\frac{r}{1-r}+\frac{r^{2}}{1-r}+\frac{r^{3}}{1-r}+\frac{r^{4}}{1-r}...\]
Which is just
\[\sum_{n=1}^{\infty}\frac{r^{n}}{1-r}\]
voila, another geometric series :)

Answer 7

Ah, amazing, thanks!

Answer 8

No problem :D

Answer 9

The rest being, of course:
\[\sum_{n=1}^{\infty}=\frac{ r^n }{ 1-r }=\frac{ 1 }{ 1-r }\sum_{n=1}^{\infty} r^n=\frac{ 1 }{ 1-r }\frac{ r }{ 1-r }=\frac{ r }{ (1-r)^2 }\]

Answer 10

That's right :D