Let $f_n:\textbf{R}\to\textbf{R}$ be $$f_n(x)=\frac{\mathrm{e}^{-n \left|x-n\right|}}{n}$$ I have to check if series $$\sum^{\infty}_{n=1}f_n(x)$$ is uniformly convergent on $\textbf{R}$.
Is the series uniformly convergent? I tried to find $M_n>\left|f_n(x)\right|$ to use Weierstrass test, but failed to find a sequence such that the inequality will hold for all $x$.
Fixed a natural number $N$. Among the terms $\frac{e^{-n|x-n|}}{n}$, $n\geq N$, some of which satisfy $|x-n|\geq 1$, in which case $\frac{e^{-n|x-n|}}{n}\leq\frac{e^{-n}}{n}$; some of which satisfy $|x-n|<1$, all the possible candidates being $n=\lfloor x\rfloor, \lfloor x\rfloor+1$ but only when $\lfloor x\rfloor\geq N$, $\lfloor x\rfloor+1\geq N$. Thus we may write $$\sum_{n\geq N}\frac{e^{-n|x-n|}}{n}= \sum_{n\geq N, |x-n|\geq 1}\frac{e^{-n|x-n|}}{n} +\sum_{n\geq N, |x-n|< 1}\frac{e^{-n|x-n|}}{n}$$ where $$\sum_{n\geq N, |x-n|\geq 1}\frac{e^{-n|x-n|}}{n}\leq \sum_{n\geq N}\frac{e^{-n}}{n}\leq \sum_{n\geq N}e^{-n}\to 0,\text{ as }N\to \infty,\text{ uniformly in }x$$ and $$\sum_{n\geq N, |x-n|< 1}\frac{e^{-n|x-n|}}{n}\leq\sum_{n=\lfloor x\rfloor, \lfloor x\rfloor+1, n\geq N}\frac{1}{n}\leq \frac{2}{N}\to 0,\text{ as }N\to\infty, \text{ uniformly in }x.$$