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.

Looking at the given series we see that uniform convergence is endangered by the fact that for arbitrary large $n$ we can find an $x$ where the $n$-th term is large, namely $\ =e^0=1$. Therefore we try to prove that the convergence is not uniform by exploiting this fact.
Put $s_n(x):=\sum_{k=0}^n e^{-|x-n|}$. By Cauchy's criterion uniform convergence means that for any $\epsilon>0$ there is an $n_0$, depending on $\epsilon$, such that
$$\bigl|s_m(x)-s_n(x)\bigr|<\epsilon\qquad\forall m>n_0,\ \forall n>n_0,\ \forall x\in{\mathbb R}\ .$$
Assume now there is such an $n_0$ for $\epsilon:={1\over2}$ and put 
$$n:=n_0+1, \quad m:=n_0+2, \quad x:=m\ .$$
Then $s_n$ and $s_m$ differ by just one term (the $m$th), and one has
$$\bigl|s_m(x)-s_n(x)\bigr|=e^{-|x-m|}=e^0>{1\over2}=\epsilon\ ,$$
which is a contradiction.
