Example of uniform but not normal convergence Let's consider a function series defined as follows.
For every $n\in \mathbb N, n\gt 1$, let $ f_n(x):= \frac{1} {n} $ if $x=\frac{1} {n}$ and $f_n(x)=0$ $x\neq\frac{1} {n}$.
Prove, using the definition of uniform convergence and the Weierstrass M-test, that $$\sum_{n=1}^{+\infty }f_n(x) $$ converges uniformly but not normally in $\mathbb R$.
(I assume a series is normally convergent if $\sum_{n=1}^{+\infty } sup \{ \vert f_n(x) \vert \} $ converges)
My attempt.
For every $n$, $sup \{ \vert f_n(x) \vert; x \in \mathbb R \}= \frac{1} {n} $ and the harmonic series diverges. So we don't have normal convergence.
Though, I'm not sure how to do with the uniform convergence. It may seem a trivial question, but I got lost with the epsilons and the supremums. I suspect that it converges to the 0 function, but $ sup \{ \vert \sum_{k=1}^{n} f_k(x) \vert; x \in \mathbb R\} \ge 1=f_1(1)=\sum_{k=1}^{n} f_k(1)$, so it does not tend to 0.
Any help, using the definition or the epsilon?
 A: Given $\epsilon > 0$, note that with $m > n > N$ and $\frac{1}{N} < \epsilon$
$$\left|\sum_{j=n+1}^m f_j(x) \right| = \begin{cases}0, &x \neq \frac{1}{p} \, \text{ where }\,\, p \in \mathbb{N},\\ 0, &x = \frac{1}{p}\, \text{ where }\,\, p \in \mathbb{N},  p \not\in(n,m]\\ \frac{1}p, & x= \frac{1}{p} \, \text{ where }\,\, p \in \mathbb{N}, n < p \leqslant m \end{cases} $$
Since $\frac{1}{p} < \frac{1}{n} < \frac{1}{N} < \epsilon$, it follows that for all $m > n > N$ and all $x \in \mathbb{R}$,
$$\left|\sum_{j=n+1}^m f_j(x) \right| < \epsilon$$
and we have uniform convergence of the series by the Cauchy criterion.
A: My initial attempt mixed up.  I was looking at the right sum, but drew the wrong conclusion.  I must not be feeling well today, because it's the second answer I've screwed up.
Let $$S=\left\{1/n\mid n\in \mathbb{Z}^+\right\}$$  Then it's obvious that the sum converges pointwise to $$f(x)=\begin{cases}x,&x\in S\\0,&\text{otherwise}\end{cases}$$ because for each $x$, $f_k(x)\neq0$ for at most one value of $k$.  Then we have $$f(x)-\sum_{k=1}^nf_k(x)=\begin{cases}x,&x\in S, \frac1x>n\\0,&\text{otherwise}\end{cases}$$
For all $n>\frac1\varepsilon$, the above difference is $<\varepsilon$, so the convergence is uniform.
In answer to your comment, note that
$$\left\lvert f(x)-\sum_{k=1}^nf_k(x)\right\rvert=\left\lvert\sum_{k=n+1}^nf_k(x)\right\rvert$$ and uniform convergence means that the last expression is uniformly small.
