# A uniformly but not normally convergent function series

Consider a functions series $$\displaystyle \sum f_n$$ of functions defined on a set $$S$$ to $$\mathbb R$$ or $$\mathbb C$$. It is known that if $$\displaystyle \sum f_n$$ is normally convergent, then $$\displaystyle \sum f_n$$ is uniformly convergent.

The converse is not true and we provide two counterexamples.

Consider first the sequence of functions $$(g_n)$$ defined on $$\mathbb R$$ by:
$g_n(x) = \begin{cases} \frac{\sin^2 x}{n} & \text{for } x \in (n \pi, (n+1) \pi)\\ 0 & \text{else} \end{cases}$ The series $$\displaystyle \sum \Vert g_n \Vert_\infty$$ diverges as for all $$n \in \mathbb N$$, $$\Vert g_n \Vert_\infty = \frac{1}{n}$$ and the harmonic series $$\sum \frac{1}{n}$$ diverges. However the series $$\displaystyle \sum g_n$$ converges uniformly as for $$x \in \mathbb R$$ the sum $$\displaystyle \sum g_n(x)$$ is having only one term and $\vert R_n(x) \vert = \left\vert \sum_{k=n+1}^\infty g_k(x) \right\vert \le \frac{1}{n+1}$

For our second example, we consider the sequence of functions $$(f_n)$$ defined on $$[0,1]$$ by $$f_n(x) = (-1)^n \frac{x^n}{n}$$. For $$x \in [0,1]$$ $$\displaystyle \sum (-1)^n \frac{x^n}{n}$$ is an alternating series whose absolute value of the terms converge to $$0$$ monotonically. According to Leibniz test, $$\displaystyle \sum (-1)^n \frac{x^n}{n}$$ is well defined and we can apply the classical inequality $\displaystyle \left\vert \sum_{k=1}^\infty (-1)^k \frac{x^k}{k} – \sum_{k=1}^m (-1)^k \frac{x^k}{k} \right\vert \le \frac{x^{m+1}}{m+1} \le \frac{1}{m+1}$ for $$m \ge 1$$. Which proves that $$\displaystyle \sum (-1)^n \frac{x^n}{n}$$ converges uniformly on $$[0,1]$$.

However the convergence is not normal as $$\sup\limits_{x \in [0,1]} \frac{x^n}{n} = \frac{1}{n}$$.