I am reading Bak's Complex Analysis for my undergraduate complex analysis class, and I am a little confused on the line of reasoning in the proof that is given for the $M$-Test for uniform convergence. Here is their statement of the $M$-Test:
Suppose $f_k$ is continuous in $D$, $k=1,2,\dots$. If $|f_k(z)|\leq M_k$ throughout $D$ and if $\sum_{k=1}^\infty M_k$ converges, then $\sum_{k=1}^\infty f_k(z)$ converges to a function $f$ which is continuous.
And then the proof that they give is this:
The convergence of $\sum_{k=1}^\infty f_k(z)$ is immediate. Moreover, for each $\epsilon>0$, we can choose $N$ so that $$\left|f(z)-\sum_{k=1}^\infty f_k(z)\right|=\left|\sum_{n+1}^\infty f_k(z)\right|\leq M_{n+1}+M_{n+2}+\dots<\epsilon$$ for $n\geq N$. Hence the convergence is uniform and $f$ is continuous.
So the first equality makes sense, and the $\leq$ makes sense since we are assuming that $f_k\leq M_k$ for all $k$. What confuses me is how we can say that $M_{n+1}+M_{n+2}+\dots<\epsilon$. I think it comes from the definition of convergence, but it's not immediately clear to me, so any insight would be appreciated.
Since the series $\sum_{k=1}^\infty M_k$ converges,$$\lim_{n\to\infty}\sum_{k=n+1}^\infty M_k=\lim_{n\to\infty}\left(\sum_{k=1}^\infty M_k-\sum_{k=1}^nM_k\right)=0.$$So, given $\varepsilon>0$, we have $\sum_{k=n+1}^\infty M_k<\varepsilon$ if $n$ is large enough.