$(f_n):[0,\infty[ \rightarrow \mathbb{R}$ : $$\forall n \in \mathbb{N}, \: \: f_n(t)=\frac{1}{n+1}\chi_{[n,n+1[}$$
where $\chi_{[n,n+1[}$ is the indicator function
this problem is confusing for me, how can i study the convergence of such series? if you have more problems of this type please let me know so i can work on them, thanks for reading. (sorry for the poor format)
For fixed $t\ge0$ you have that $f_n(t)=0$ for all $n$ except, possibly, two of them (this follos from the fact that the supports of $f_n$ are $[n,n+1]$). So the pointwise, it is a finit sum, so convergent.
For the uiniform convergence, observe that you cannot use M-Weierstrass test, since $\sup f_n(t)=\frac{1}{n+1}$ and $\sum\frac{1}{n+1}$ diverges.
The solution is to use uniform-Cauchy-test for uniform convergence.
For this purpose, let $n>m$. For fixed $t\ge 0$ if you take $$f_{m+1} (t)+\cdots+f_n(t)$$ there are AT MOST two of these sumands that are non zero, so for any $t\ge 0$ $$0\le f_{m+1}(t)+\cdots+f_n(t)\le 2\frac{1}{m+1}$$ (take into account that $\frac{1}{m+1}$ is the biggest possible value).
So Given $\varepsilon>0$ if fou take $N\in\Bbb N$ such that $\frac{2}{N+1}<\varepsilon$, for every $n>m\ge N$ fou have that for all $t\ge 0$, $$0\le \sum_{k=m+1}^n f_k(t)\le \frac{2}{m+1}\le \frac{2}{N+1}\le\varepsilon$$ and you have that your series is unifomly Cauchy, so uniformly convergent.
PS: The only property I'm strongly using is that the supports are "almost" disjoint (i.e. the intersection two different supports is empty os a singleton).
I have recently co-authores a paper very related to this topic: https://arxiv.org/abs/2003.10263 where you can find more examples of this type.