Convergence of $\sum_{n=0}^{\infty}ne^{-\beta n}$

Question

I am working in a problem of statistical mechanics and a partition function I found is of the form: $$ Z = \sum_{n=0}^{\infty}(n+1)e^{-\beta n} $$ I used the ratio test and the series converges. I want to find a closed form for this. I tried: $$ \sum_{n=0}^{\infty}(n+1)e^{-\beta n} = \sum_{n=0}^{\infty}ne^{-\beta n} + \sum_{n=0}^{\infty}e^{-\beta n} $$ The second term of the right-hand side is simply $1/(1-e^{\beta})$ (geometric series). So the question comes down to: does this series have a closed form? $$ \sum_{n=0}^{\infty}ne^{-\beta n} $$ where $\beta>0$

Answer 1

Hint

Yes. Denote $f(\beta)$ the geometric series $$f(\beta)=\sum_{n=0}^\infty e^{-\beta n}$$ what's $f'(\beta)$? Can you take it from here?

Answer 2

remember the expressions

$$ \frac{x}{1+x}= \sum_{n=1}^{\infty}x^{n} $$

$$ \frac{1}{1+x}= \sum_{n=0}^{\infty}x^{n} $$