I am trying to understand a proof of Weierstrass M-test stated as
"If $|u_k(x)|\leq a_k$ for all $x\in M$ and $\sum_{k=1}^\infty a_k$ converges, then $\sum_{k=1}^\infty u_k(x)$ konverges uniformly on $M$."
My attempt to prove the theorem goes like this:
Since $\sum_{k=1}^\infty a_k$ converges there is to every number $\epsilon>0$ a number $N>0$ such that whenever $n\geq N$ we have
$$|\sum_{k=1}^\infty a_k - \sum_{k=1}^n a_k| = |\sum_{k=n+1}^\infty a_k| = \sum_{k=n+1}^\infty a_k < \epsilon$$
Thus, for $n\geq N$,
$$|\sum_{k=1}^\infty u_k(x) - \sum_{k=1}^n u_k(x)| = |\sum_{k=n+1}^\infty u_k(x)| \leq \sum_{k=n+1}^\infty |u_k(x)| \leq \sum_{k=n+1}^\infty a_k < \epsilon$$
for all $x\in M$.
This was my attempt. However, the proof I am looking at wants to first show that $\sum_{k=1}^\infty u_k(x)$ converges pointwise to a sum $s(x)$, and then show that when $n\geq N$ then $|s(x) - s_n(x)| < \epsilon$. Is this really necessary?
Your proof is wrong, since you mention $\sum_{k=1}^\infty u_k(x)$ without having proved that this series converges.