# Finding infinite series formulae

1. Jun 18, 2012

### Contingency

1. The problem statement, all variables and given/known data
This isn't my homework problem, but I'd like to know how to prove $$\sum _{ n=0 }^{ \infty }{ \frac { n }{ { \alpha }^{ n } } } =\quad \frac { \alpha }{ { (\alpha -1) }^{ 2 } }$$

I only know basic convergence tests (including the integral test), and that
$$\forall |\alpha |<1\quad \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \alpha }^{ n } } } =\frac { 1 }{ 1-\alpha }$$

Also, can anyone list some tools to tackle these problems with?

2. Relevant equations
$$\forall |\alpha |<1\quad \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \alpha }^{ n } } } =\frac { 1 }{ 1-\alpha }$$

Edit: correction to geometric series equation: $$\forall |\alpha |<1\quad \sum _{ n=0 }^{ \infty }{ { \alpha }^{ n } } =\frac { 1 }{ 1-\alpha }$$

Last edited: Jun 18, 2012
2. Jun 18, 2012

### LCKurtz

Certainly, that equation isn't correct; try $\alpha = \frac 1 2$. But to give a general answer to your question, when you have a sum similar to $$\sum_{n=0}^\infty nx^n$$which would be a geometric series if that $n$ weren't out in front, you want to think about derivatives. Start with$$f(x) = \sum_{n=0}^\infty x^{n} = \frac 1 {1-x}\hbox{ if }|x|<1$$Differentiate both sides of that:$$\frac 1 {(1-x)^2}= \sum_{n=0}^\infty nx^{n-1} = \frac 1 x\sum_{n=0}^\infty nx^n$$
Multiply both sides by $x$ and you have$$\sum_{n=0}^\infty nx^n = \frac x {(1-x)^2}$$Then put in whatever $x$ you wish, as long as $|x|<1$.

Last edited: Jun 18, 2012
3. Jun 18, 2012

### clamtrox

For starters, your geometric series is not right... It should be $\alpha^n$, not $1/\alpha^n$.

This looks a lot like you might want to differentiate the geometric series with respect to alpha. It should be easy to convince yourself that this is legal.

4. Jun 18, 2012

### Contingency

Thank you very much

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