Q:

Does the series $s_n(x)=\sum^{n}_{k=0}\frac{e^{-kx}}{1+kx}$ converge uniformly over $ [1,2]?$

Does the series \begin{align}s_n(x)=\sum^{n}_{k=0}\frac{e^{-kx}}{1+kx},\;x\in [1,2]\end{align}
converge uniformly?
MY TRIAL:
I thought, by intuition, that $s_n(x)$ does not converge uniformly, which implies that $s_n(x)$ is not uniformly Cauchy i.e., $\exists\,\epsilon_0>0$ s.t. $\forall\,k\in \Bbb{N}, \,\exists\,n_k,m_k\geq k,\;\exists\,x_k\in[1,2]$ s.t. \begin{align}\left|s_{n_k} -s_{m_k}\right|\geq \epsilon_0.\end{align}
Let $k\in \Bbb{N}$ be given. Take $n_k=k,\,m_k=k+1,\,x_k=1+\frac{1}{k+1}$ and . Then,
\begin{align}\left|s_{n_k} -s_{m_k}\right|&=\left|\sum^{n_k}_{p=0}\frac{e^{-px_k}}{1+px_k}-\sum^{m_k}_{p=0}\frac{e^{-px_k}}{1+px_k}\right|\\&=\left|\sum^{k}_{p=0}\frac{e^{-px_k}}{1+px_k}-\sum^{k+1}_{p=0}\frac{e^{-px_k}}{1+px_k}\right|\\&=\left|\frac{e^{-(k+1)\left(1+\frac{1}{k+1}\right)}}{1+(k+1)\left(1+\frac{1}{k+1}\right)}\right||\\&=\left|\frac{e^{-k+2}}{3+k}\right|\end{align}
From here, I don't know what to do. Any help please? Anyway, I don't even know if my guess was right. It could be that the series converges uniformly.

A:

Hint : Use Weierstrass-M test and the fact that $|\frac{e^{-k x}}{1+k x} |\leq \frac{e^{-k}}{1+k}$ for all $x \in [1,2]$
Edit : $ 1+k \leq 1+ k x =>  k \leq k x => 1 \leq x$ and this is true since $x \in [1,2]$, so the denominator is the smallest when $x=1$.
$ e^{-k x} \leq e^{-k} $ => applying log to both sides of the inequality gives $ -k x \leq -k$ => dividing by $-k$ and flipping the inequality (division by negative number)  to get $x \geq 1$ and this is true since $x \in [1,2]$
By Weierstrass-M test and the fact of $\sum \limits_{k=1}^{\infty} \frac{e^{-k}}{1+k}$ convergence we get that the original sum is uniformly convergent when $x \in [1,2]$.

