Prove uniform convergence and identity for the series $\sum^{\infty}_{n=0} e^{-nx}$ and $\sum^{\infty}_{n=0} ne^{-nx}$ where $x > 0$. 
Prove uniform convergence and identity for the series $\sum^{\infty}_{n=0} e^{-nx}$ and $\sum^{\infty}_{n=0} ne^{-nx}$ where $x > 0$. 

Consider $\sum^{\infty}_{n=0} e^{-nx}$ and $\sum^{\infty}_{n=0} ne^{-nx}$ where $x > 0$.
I've proved that $$\sum^{\infty}_{n=0} e^{-nx} = \frac 1 {1-e^{-x}}$$.
I must show both series for $x_0 > 0$ that both series converges uniformly in $[x_0, \infty)$. To do this I've tried to find majorant series for both but I've been unsuccesful.
$$e^{-nx} = \frac 1 {e^{nx}} \le \frac 1 {2^{nx}}\le \frac 1 {2^{nx_0}}$$, but this doesn't give a proper majorant series.
Also I must show $$\sum^{\infty}_{n=0} ne^{-nx} = \frac {e^x} {(e^x-1)^2}$$
I have $(e^{-nx})^{'} = -ne^{nx}$. Can I show $$\sum^{\infty}_{n=0} -ne^{-nx}$$ converge uniformly ? (how) - if so then $$\sum^{\infty}_{n=0} -ne^{-nx}= \frac {-e^x} {(1-e^{-x})^2}$$ which is not the result I was hoping for ?
 A: Let $a>0$ then we have
$$|e^{-nx}|\le e^{-na},\quad\forall x\ge a$$
and the series 
$$\sum_{n\ge0}e^{-na}$$
is convergent so we have the uniform convergence of $\sum_n e^{-nx}$ on every interval $[a,\infty)$ and clearly 
$$\sum_{n=0}^\infty e^{-nx}=\frac1{1-e^{-x}}$$.
By the same method and using that
$$|ne^{-nx}|\le ne^{-na},\quad \forall x\ge a$$
we prove the uniform convergence of $\sum_nne^{-nx}$ on the interval $[a,+\infty)$ and since 
$$(e^{-nx})'=-ne^{-nx}$$ so we have
$$\sum_{n=0}^\infty ne^{-nx}=-\frac{d}{dx}\left(\sum_{n=0}^\infty e^{-nx}\right)=-\frac{d}{dx}\left(\frac1{1-e^{-x}}\right)=\frac{e^{-x}}{(1-e^{-x})^2},\quad\forall x>0$$
A: You get the second one by differentiating the first w.r.t. $x$. Since you have proved that it converges uniformly, you can differentiate $\displaystyle \sum_{n=1}^{\infty} e^{-nx} = \frac{1}{1  - e^{-x}}$ term by term on the left side as well as the one on right side.
This type of sequence is called Arithmetico-Geometric sequence. You can do the the following and show that it converges uniformly. 
$$S_m(x) = \sum_{k=1}^{m} k e^{-kx} \tag{1}$$
Multiply $(1)$ by $e^{-k}$ and subtracting the $(1)$ gives, 
$$ e^{-kx}S_m(x) - S_m(x) = \sum_{k=1}^{m}ke^{-(k+1)x} - \sum_{k=1}^{m}ke^{-kx}  \\ = me^{-(m+1)x} + \sum_{k=2}^{m}(k-1)e^{-kx} - \sum_{k=2}^{m}k e^{-kx} - e^{x}\tag{2}$$
You can find the value of $S_m(x)$ from $(2)$ and show that it converges uniformly to desired value.
