Prove or disprove: $\sum_{n=0}^\infty e^{-|x-n|}$ converges uniformly in $(-\infty,\infty)$

I’m trying to find out if $\sum_{n=0}^\infty e^{-|x-n|}$ converges uniformly in $(\infty,\infty)$. Here’s my attempt at an answer:

If $ x\in(-\infty,0]:$
$$e^{-|x-n|}=e^{(-(x-n))}=e^{x-n}\le e^{-n}$$
$\sum e^{-n}$ converges, therefore $\sum_{n=0}^\infty e^{-|x-n|}$ converges uniformly in $(-\infty,0]$.

If $x\in[0,\infty):$

Here no series $\sum a_n$ exists so that $e^{-|x-n|} \le a_n$ for every $x\in[0,\infty)$, because for every n, there exists $x\space | \space e^{-|x-n|}=1$, for this reason i think that perhaps the series doesn’t converge uniformly in $(\infty,\infty)$.

Could someone show me how to continue from here?

Solutions Collecting From Web of "Prove or disprove: $\sum_{n=0}^\infty e^{-|x-n|}$ converges uniformly in $(-\infty,\infty)$"

You are right, the series doesn’t converge uniformly on $[0,+\infty)$, and you have correctly identified the reason for that.

Now all that remains is to make a formal argument of that. We get that from the

Lemma: Let $f_n \colon S \to \mathbb{R}$ be a sequence of functions such that the series $\sum\limits_{n = 0}^{\infty} f_n(x)$ is uniformly convergent on $S$. Then $(f_n)$ converges uniformly to $0$ on $S$.

Proof: Let $g \colon S \to \mathbb{R}$ be the sum of the series, and $p_k(x) = \sum_{n = 0}^k f_n(x)$ the $k^{\text{th}}$ partial sum function. Given $\varepsilon > 0$, by the uniform convergence of the series, there is a $K(\varepsilon) \in \mathbb{N}$ such that

$$\sup \{ \lvert p_k(x) – s(x)\rvert : x \in S\} \leqslant \frac{\varepsilon}{2}$$

for every $k \geqslant K(\varepsilon)$. Then, for every $k > K(\varepsilon)$ and every $x\in S$ we have

\begin{align}
\lvert f_k(x)\rvert &= \lvert p_k(x) – p_{k-1}(x)\rvert = \bigl\lvert \bigl(p_k(x) – s(x)\bigr) + \bigl(s(x) – p_{k-1}(x)\bigr)\bigr\rvert\\
&\leqslant \lvert p_k(x) – s(x)\rvert + \lvert p_{k-1}(x) – s(x)\rvert \leqslant \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon,
\end{align}

whence $\sup \{ \lvert f_k(x)\rvert : x\in S\} \leqslant \varepsilon$ for $k > K(\varepsilon)$, and the uniform convergence of $(f_n)$ to $0$ on $S$ is proved.

In your example, with $f_n(x) = e^{- \lvert x-n\rvert}$ on $\mathbb{R}$, as you observed we have $f_n(n) = 1$ for all $n$, and so the sequence $(f_n)$ doesn’t converge to $0$ uniformly on $\mathbb{R}$, whence by the lemma the series cannot converge uniformly on all of $\mathbb{R}$.

But, with a minuscule modification, your argument for the uniform convergence of the series on $(-\infty, 0]$ shows that the series converges uniformly on $(-\infty, A]$ for every $A\in \mathbb{R}$.

This is not an answer, just an illustration. Here is a graph of $\phi(x) = \sum_{n = -\infty}^\infty e^{-|x-n|}$ for $-5 \le x \le 5$.

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