Show $\sum n e^{-na}$ converges for $a>0$ Is there any test or property in particular I can use to show $
\sum n e^{-n a}$ is convergent for $a>0$ ?
I think it is obvious that from looking at the function that this is convergent, since 
$n e^{-na} \to 0$ as $n \to \infty$ since the exponential out-decays the linear term.
Although, the harmonic series comes to mind, and we all know $\sum \frac{1}{n}$ is not convergent even though $\frac{1}{n} \to 0$ as $n \to \infty$
In the case of my question, there is an upper bound: if we take $C \in \mathbb{R} $ then $n e^{-na} \leq C e^{-n \frac{a}{2}}$
Where can I go from here to construct a more rigorous proof?
Would I just need to show $\sum e^{-n \frac{a}{2}}$ converges?
 A: I think that the best way is to use the root test:http://en.wikipedia.org/wiki/Convergence_tests. Indeed you have $(ne^{-an})^{\frac{1}{n}}\rightarrow e^{-a}<1$.
A: Another easy test you can use: the ratio or d'Alembert test.
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-(n+1)a}}{ne^{-an}}=\frac{n+1}n\cdot\frac1{e^a}\xrightarrow[n\to\infty]{}\frac1{e^a}<1$$
A: You can also use a comparison test by noticing that $ne^{-na}=O\left(\frac{1}{n^2}\right)$ for $a>0$.
A: we know that sigma n x^n converges to x/(1-x)^2 for abs(x)<1. This is easy to show by ratio or root test. our series is the same as sigma n(1/e^a)^n,0<1/e^a<1, since a>0. In fact we can fond the sum.
A: What if you define $x=e^{-a}$ ? This would make $$S(a)=\sum_{n=1}^\infty n e^{-n a}=\sum_{n=1}^\infty n x^n=x \sum_{n=1}^\infty n x^{n-1}=x \frac{d}{dx}\Big( \sum_{n=1}^\infty x^n\Big)$$ You have then a classical summation; differentiate it to get $$S(a)=\frac{x}{(1-x)^2}=\frac{e^a}{\left(1-e^a\right)^2}$$ which undefined only for $a=0$; you could also note that $S(-a)=S(a)$.
