I have been fiddling around with some series of functions and analyzing whether they converge pointwise or uniformly. Furthermore I know that continuity and convergence of integrals does not always follow from pointwise but for uniform convergence as seen in a counterexample (of a non-uniform convergence) for $f_n:[0,1]\to\mathbb R$ with
$$f_n(x)=\begin{cases}n^2x, &0\leq x\leq \frac1n,\\2n-n^2x, &\frac1n<x\leq \frac2n,\\ 0, &x>\frac2n,\end{cases}$$
which yields $\lim_{n\to\infty}f_n(x)=0$ for all $x\in[0,1]$ but $\int_0^1f_n(x)~\mathrm dx=1\neq 0$.
I am having trouble finding a decent informal explanation (not just applying the definitions to test for convergence) of both terms other than referring to the "speed of convergence" which is different in both cases.
ELI5: What are pointwise and uniform convergence and what is the difference?
We can start from the definitions. For a sequence $f_n(x)$ with $f_n: S\to \mathbb{R}$, $n\in \mathbb{N}$ we have:
and
Note that the two quantifiers $\{\exists N \in \mathbb{N}\}$ and $\{\forall x\in S\}$ change position in these two definitions, and this is the key difference between them.
The number $N$ is the thing that exactly define the ''speed of convergence''. For the uniform convergence this speed is fixed and it is the same for all $x$, For the pointwise convergence this speed can change (also dramatically) for different $x$.
This is the case of your example. The functions in your sequence have a spike at $x=1/n$ whose base becomes little with $n$ increasing, but whose height becomes higher. See the figure where there are represented the functions for $n=4$ and $n=5$.
You can see that the number $N$ becomes more an more great as $x\to 0$ because we have to limit the base of the spike at a value $<x-\epsilon$.
For the uniform convergence the situation is quite different because , for a given $\epsilon$ the number $N$ is the same at any value of $x$, as you can see in this image that I've found at: https://simomaths.wordpress.com/2012/12/23/basic-analysis-uniform-convergence/.