Does $\sum_{n=0}^{\infty}e^{-|x-n|}$ converge and uniformly converge? For $ 0 \leq x < \infty$, I'd like your help with deciding whether the following series converges and uniformly converges: $\sum_{n=0}^{\infty}e^{-|x-n|}$.
$$\sum_{n=0}^{\infty}\frac{1}{e^{|x-n|}}=\sum_{n=0}^{N}\frac{1}{e^{|x-n|}}+\sum_{i=0}^{\infty}\frac{1}{e^{i}}.$$
I divided the sum into two sums where $N$ is the place where $x=n$, so the first sum is finite and the second one converges, so the total sum converges. Am I allowed to say that the sum is uniformly converges, since the second series, where the tail converges, does not depend on $x$? 
Thanks.

We can show that the series is uniformly convergent on bounded subsets of $\mathbb R_+$. Indeed, let $B$ such a set and $R$ an integer such that $B\subset [0,R]$. If $n>R$ then $n-x>R-x\geq 0$ os $|n-x|=n-x$ and $0\leq e^{-|n-x|}=e^{-(n-x)}=e^{-n}e^x\leq e^{-n}e^R$, so in fact the series $\sum_{n\geq 0}e^{-|n-x|}$ is normally convergent on $[0,R]$ hence on $B$. 
If $U$ is an unbounded subsets of $\mathbb R_+$, put $I_k:=[k,k+1[$. Let $n_k$ an increasing sequence of integers such that $I_{n_k}\cap U\neq\emptyset$. Then for each $n\in \mathbb N$, and $k'$ such that $n< n_{k'}$
$$\sup_{x\in U}\sum_{k=n}^{+\infty}e^{-|k-x|}\geq \sup_{x\in U}\sum_{j=k'}^{+\infty}e^{-|n_j-x|}\geq \sup_{x\in U\cap I_{n_{k'}}}e^{-|n_{k'}-x|},$$
and for $x\in U\cap I_{n_{k'}}$ we have $-|n_{k'}-x|=-x+n_{k'}\geq -1$ so 
$\sup_{x\in U}\sum_{k=n}^{+\infty}e^{-|k-x|}\geq e^{-1}$
which shows that the convergence cannot be uniform on $U$.
