Convergence of Series of Functions $\sum_{n = 1}^{\infty}(n + 1)e^{1 - nx}$ I'm learning about series of functions and need some help with this problem: 

Given the series of function $\sum_{n = 1}^{\infty}(n +  1)e^{1 - nx}$ show that 
(i) it converges pointwise but not uniformly on the interval $(0, +\infty)$; 
(ii) it converges uniformly on the interval $(1, +\infty)$. 

My work and thoughts:
Since I'm having difficulties showing (i) I'll be explaining my work for (ii).
(ii) We note that $\sum_{n = 1}^{\infty}(n +  1)e^{1 - nx} = 
e \sum_{n = 1}^{\infty}\frac{(n +  1)}{e^{nx}}$ and let  $f_n(x) = 
\frac{(n + 1)}{e^{nx}}$. 
Therefore $f'_n(x) = \frac{-(n + 1)ne^{nx}}{e^{2nx}} =  
\frac{-(n + 1)n}{e^{nx}} < 0 \ \forall{x} \in (1, +\infty)$.
So $f_n$ is decreasing on the interval $(1, +\infty)$. In other words $f_n$ is bounded from above and we can write 
$$\forall{x} \in (1, +\infty) : |f_n(x)| \leq f_n(1) = \frac{n + 1}{e^{n}}.$$
It is easy to prove that the series 
$\sum_{n = 1}^{\infty} \frac{n + 1}{e^{n}} < +\infty$
(the series converges by the Limit Comparaison Test). 
Hence, by the Weierstrass M-test, we conclude that the given series $\sum_{n = 1}^{\infty}(n +  1)e^{1 - nx}$ is uniformly convergent on the interval 
$(1, +\infty)$. 

Is my work correct for (ii)? How do I show that (i) the series of functions converges pointwise but not uniformly on the interval $(0, +\infty)$? 
 A: Note that
$$
\sum (n+1)q^n=\frac1{(1-q)^2}
$$
and convergence is uniform for $|q|\le r$ for any $r<1$.
A: Yes, what you did for (ii) is correct. 
For (i), the pointwise convergence follows from the ratio test.
If the series would be uniformly convergent on $(0,+\infty)$, then the sequence $\left(f_n\right)_{n\geqslant 1}$ would converge uniformly on this interval. This would imply that $\lim_{n\to +\infty}f_n(1/n)=0$.
A: Let $$\:F_n(x)=\sum_{j=1}^{n}f_j(x)\:, \:F(x)=\sum_{j\in\mathbf N}f_j(x),\quad\:f_j(x)=(j+1)e^{1-jx}\:\:\:\&\:\:\:(j,n)\in\mathbf N^2\:\:\:\forall j\le n.$$
We suspect that $$\exists\hat{\large\epsilon}>0\:\:\forall N\in\mathbf N,\:\:\exists(k,n)\in\mathbf N^2\:\:\text{ such that for }\:n\ge k\ge N\implies\sup_{0<x\le1}\left|\sum_{j=k}^nf_j(x)\right|=\:...\\...=\left\|F_n(x)-F_k(x)\right\|_{0<x\le1,\infty}>\large \hat\epsilon.$$
For any $\text{fixed }\:N\in\mathbf N\:$ if we choose $\:k=2^N,n=2^{N+1}-1\:$ then we have :
$$\left\|F_n(x)-F_k(x)\right\|_{0<x\le1,\infty}=\sum_{j=2^N}^{2^{N+1}-1}e(j+1)>e\:{(2^{N+1}-1)2^{N+1}-2^N(2^N+1)\over 2}>6=\large \hat\epsilon.$$
By the Cauchy criterion for series of functions, $\:F(x)\:$ fails to converge uniformly on $\:(0,1],\:$ hence the uniform convergence is not applicable on $\:(0,\infty).$
