# If sequence of function does not converges uniformly, so doesn't the series?

It's been more than a year since I took a course in which we study sequences and series of functions.

In some notes I read (without proof) that if a sequence of functions $f_n$ doesn't converges uniformly (in some interval), then the series of functions $\sum f_n$ so does neither. This might be just a trivial fact that I am not seeing.

I mean, for regular series $\sum a_n$ we always need $\lim a_n=0$ or the series will diverges. Similarly, we need $\lim |f_n-f|_A=0$ to the sequence of functions $f_n$ to converges uniformly to $f$ in $A$, so it seems reasonable that this be a necessary condition to the series $\sum f_n$ to converge uniformly.

So, I am asking, can this be use as a criterion of non-uniform convergence of $\sum f_n$? How can we prove this?

• If $\sum f_n$ uniformly converges, then $f_n \to 0$ uniformly. – user384138 Jan 20 '17 at 17:55
• yup, and the converse is false, it's pretty much the same thing as in calc 1 – Jorge Fernández Hidalgo Jan 20 '17 at 17:56

You have the conditional statement:

$\sum f_n$converges uniformly $\implies f_n \rightarrow 0$ uniformly

Think about the contrapositive statement.

Suppose that $\sum\limits_{n=1}^\infty f_n$ converges to $f$. This means that for each $\epsilon$ there is an $N$ such that $\sum\limits_{n=1}^m |f_n(x)-f(x)|<\epsilon$ for all $m\geq N$ and $x\in X$.

in particular we must have $|\sum\limits_{n=1}^{m+1}f_n(x)-\sum\limits_{n=1}^{m} f(x)|<2\epsilon$ for all $x$ by the triangle inequality.

so $|f_{m+1}(x)|<2\epsilon$ for all $m\geq N$ and $x\in X$.

Clearly we can use this to prove $f_n$ converges uniformly to the constant $0$ function taking $N'=N+1$ and choosing $\epsilon$ to be half of a given $\epsilon'$.