The Weierstrass M-test tells that given a function sequence $(u_{n}(x))$ where $x \in I$, if there exists a convergent series $\sum a_{n}$ such that $|u_{n}(x)|\leq a_{n}$ for all $n$ and $x\in I$, then $\sum u_{n}(x)$ converges uniformly in $I$.
What about the opposite of it?
If $\sum u_{n}(x)$ converges uniformly in $I$, then there exists a convergent series $\sum g_{n}$ such that $|u_{n}(x)|\leq g_{n}$ for all $n$ and $x \in I$.
Since the theorem isn't in the form of 'if and only if', I'm trying to think of an example to counter the above:
My attempt was to define a function sequence such as this:
Then taking $u_{1}(x)=f_{1}(x),\ u_{n}(x)=f_{n}(x)-f_{n-1}(x)$. It can be shown that $f_{n}(x)$ converges uniformly in $[1, \infty)$, therefore $u_{n}(x)$ does as well.
But $|u_{n}(1)|=|f_{n}(1)|$ which is the constant sequence: $1, 1, 1, ...$. If we assume by contradiction that there exists such a sequence $1\leq g_{n}$ that $\sum g_{n}$ converges then by the comparison test $\sum 1$ converges which is obviously not true.
First, I'd like to know if the above is true. It took me quite a while to come up with something and I'm not even sure it's true.
Also, It's very difficult for me to visualize these complicated functions (like the one above) where both $x$ and $n$ play a role. Is there an easier way to deal with these questions?