Show that a series of functions converges (uniformly)

Question

Let $(f_n)_{n \geq1}$ be a sequence of functions defined by $f_n: \mathbb{R}_0 \to \mathbb{R}: x \mapsto \begin{cases} \frac{1}{x} \quad \quad x \in[n-1,n) \setminus\{0\}\\0 \quad \quad \mathrm{else}\end{cases}$

Does $\sum_{n=1}^\infty f_n$ converge? Does it converge uniformly?

For $x \in \mathbb{R}_0: f_n(x) = \begin{cases} 1/x \quad \quad n = \lfloor{x+1}\rfloor \\0\quad\quad \mathrm{else}\end{cases}$

Hence, for every fixed $x$, the series the sequence $f_n(x)$ is only non zero in one point, and hence the series $\sum_{n}f_n(x)$ is convergent in any point $x$. Hence, the series of functions $\sum f_n$ is pointwise convergent.

However, I have trouble to prove it is (not) uniformly convergent. Any ideas?

Answer 1

HINT: set $g_n(x):=\sum_{k=1}^nf_k(x)$. Now observe that $f(x):=\sum_{k=1}^\infty f_k(x)=1/x$, that is, $(g_k(x))_{k\in\Bbb N}\to f(x)$ pointwise.

Also observe that $$\|g_n-f\|_\infty=\sup_{x\in(0,\infty)}\left|\sum_{k=1}^nf_k(x)-f(x)\right|=\sup_{x\in(0,\infty)}\left|\sum_{k=n+1}^\infty f_k(x)\right|=1/n$$