# $\sum_{n=1}^\infty u_n$ converges uniformly on $[a,b]$, but $\sum_{n=1}^\infty |u_n|$ doesn't

Find an example of a function sum $\sum_{n=1}^\infty u_n$ that converges uniformly on $[a,b]$ and converges absolutely pointwise, but $\sum_{n=1}^\infty |u_n|$ doesn't converge uniformly on $[a,b]$.

This seems pretty straightfoawrd but it's really hard to find an example. The choice of $u_n$ must be very specific. You can actually prove any power series wont work (due to Abel's theorem)

In addition, in order to prove the uniform convergence, $u_n$ must not satisfy Weierstrass' M-Test (because then $\sum |u_n|$ would converge uniformly aswell)

Take $u_n(x) = (-x)^n(1-x)$ on $[0,1]$.
Then we have that $\sum_{n=1}^{\infty} \big |(-1)^nx^n(1-x)\big| = \sum_{n=1}^{\infty} x^n(1-x) = \begin{cases} x, & x \not = 1 \\ 0, & x = 1 \end{cases}$. Obviously the convergence can't be uniform, as the sum consists of continuous functions, while the function to which it converges isn't continiuous.
On the other side $\sum_{n=1}^{\infty} (-1)^nx^n(1-x)$ is uniformly convergent as $|u_n(x)| \ge |u_{n+1}(x)|$ and $\lim_{n \to \infty} |u_n(x)| = 0$ on $[0,1]$
• Why is $u_n$ monotonic? $u_n-u_{n+1}=(-1)^n (1-x)(1+x)x^n$ – Theorem Feb 11 '18 at 19:43
• @Theorem In fact the absolute value should be decreasing. The theorem is $\sum (-1)^nu_n(x)$ is uniformly convergent if $u_n(x) \ge u_{n+1}(x) \ge 0$ and $\lim_{n \to \infty} u_n(x) = 0$ – Stefan4024 Feb 11 '18 at 19:44