Proving that infinite sum of negligible functions might be not negligible

Question

I have to prove that infinite sum of negligible functions might be not negligible. How can I prove that?

I know that a finite sum of negligible functions must be negligible.

thank you

Answer 1

Folowing your def'n in your comment that $f$ is negligible iff for all $k \in \mathbb N$ we have $f(n)=o(n^{-k})$ as $n\to \infty.$

(1). If $f_m(x)=e^{-x}$ for each $m\in \mathbb N$ then each $f$ is negligible but $\sum_mf_m(x)=\infty.$

(2). Let $[x]$ be the largest integer not exceeding $x.$ Let $f_m(x)=1$ for $x< m+1 $ and $f_m(x)=0$ for $x\geq m+1.$ Then each $f_m$ is negligible but $\sum_{m=1}^{\infty}f_m(x)= [x]$ (for $x\geq 1$), which is not negligible.

Answer 2

It's even worse than that: Every function whatsoever can be the (pointwise) sum of countably many negligible functions.

Let $f:\mathbb R\to\mathbb R_{\ge 0}$ be arbitrary, and define $$ f_n(x) = \begin{cases} f(x) & \text{if }n\le |x|<n+1 \\ 0 & \text{otherwise} \end{cases} $$ Then each $f_n$ is negligible, but $$ \sum_{n=0}^\infty f_n(x) = f(x) $$

These $f_n$ are generally discontinuous, but with a bit more ingenuity you can get each $f_n$ to be continuous, Lipschitz, smooth, etc. provided that $f$ is already similarly nice.

If you want $\sum f_n$ to converge uniformly, then you need to require that $f(x)\to 0$ for $x\to\infty$ (but this can be as slow as you like). Then the above construction will still work. On the other hand, it is easy to see that an uniformly converging sum of negligible functions will necessarily satisfy $f(x)\to 0$.