# Prove $\sum_{n=1}^\infty f_n$ converges uniformly if and only if $\sum_{n = M}^\infty f_n$ converges uniformly

Let $$f_n$$ be a sequence of functions on $$A$$. Prove that $$\displaystyle\sum_{n=1}^\infty f_n$$ converges uniformly on $$A$$ if and only if $$\displaystyle\sum_{n = M}^\infty f_n$$ converges uniformly on $$A$$ for for any $$M \in \mathbb{Z}^+$$.

We let $$F_N$$ denote the partial sum $$\sum_{n=1}^N f_n$$. Let $$M > 0$$. Then we can write $$$$\displaystyle\sum_{n=M}^K f_n = \sum_{n=1}^K f_n - \sum_{n=1}^{M-1} f_n = F_K - F_{M-1}$$$$

By definition, we say that $$\sum_{n=1}^\infty f_n$$ converges uniformly on $$A$$ if for any $$\varepsilon > 0$$, we can find $$N > 0$$ such that for all $$x \in A$$ and for all $$m,n > N$$, $$$$|F_m - F_n|=\left|\sum_{i=n+1}^m f_n \right| < \varepsilon$$$$

Here is where I'm stuck, as I'm not very sure how to understand uniform convergence of $$\sum_{n=M}^\infty f_n$$ in terms of its definition. Wouldn't it be the same as the above?

Any clarifications/hints is appreciated!

Assume that $$\displaystyle\sum_{n=1}^\infty f_n$$ converges uniformly on $$A$$ to a function $$f$$. If $$M\in\Bbb N$$, then $$\displaystyle\sum_{n=M}^\infty f_n$$ converges uniformly to $$\displaystyle f-\sum_{n=1}^{M-1}f_n$$, since, for each $$x\in A$$,$$\left|f(x)-\sum_{n=1}^\infty f_n\right|=\left|f(x)-\sum_{n=1}^{M-1}f_n(x)-\sum_{n=M}^\infty f_n(x)\right|.\tag1$$So, if the LHS of $$(1)$$ is smaller that $$\varepsilon$$, then so is the RHS and vice-versa.
The same argument shows that, if $$\displaystyle\sum_{n=M}^\infty f_n$$ converges uniformly to $$g$$, then $$\displaystyle\sum_{n=1}^\infty f_n$$ converges uniformly to $$\displaystyle g+\sum_{n=1}^{M-1}f_n$$.