Calculate $\sum_{n=1}^\infty \dfrac{nx}{e \ ^ {nx}}$

Question

I need to calculate $\sum_{n=1}^\infty \dfrac{nx}{e \ ^ {nx}}$ for $x \ge0 $.

Thanks !

Answer 1

Hint: $$\dfrac{d}{dt} \sum_{n=0}^\infty t^n = \sum_{n=1}^\infty n t^{n-1}$$ for $|t|<1$.

Answer 2

The RHS of your first inequlity should probably be $ 1\over \exp 1$ rather than $n\over \exp n$. By the way, let $q= \exp -x$, the general term of your series is $xnq^n$, so it converges either if $x=0$ or if $0\leq q <1$, i.e. $x>0$. You probably know that $\sum _{k=1} ^\infty kq^k= {q\over (1-q)^2}$, and $\sum _{k=1}^\infty {kx \over \exp kx}={x \exp -x \over (1-\exp -x)^2}$