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 \begin{equation} \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} \end{equation}

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 $m,n > N$, \begin{equation} |F_m - F_n|=\left|\sum_{i=n+1}^m f_n \right| < \varepsilon \end{equation}

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!


1 Answer 1


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 $\displaystyle g+\sum_{n=1}^{M-1}f_n$.


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