I have been attempting quite a few uniform convergence and point wise convergence questions to help prepare myself for an exam coming up in April. I haven't been getting to far on these particular questions, and frankly, I'm quite lost. Any explanations and tips on how to solve the following question would be greatly appreciated.
For this question, we consider the following series: $\sum \frac{x^n}{n2^n}$ for $n>-1$
(1) Show that the series converges point wise only on $[-2,2)$. (should I break up the interval into $[-2,0]$ , $[0,2)$ ?)
(2) Show that the series converges uniformly on $(a,b)$ for all $a$ , $b$ with $-2<a<b<2$ (I'm assuming I should use the Weierstrass M-test.)
thank you in advance!
Hints: For the first part, apply the root or ratio test. Let's use root for no particular reason: $$\bigg|\frac{x^n}{n2^n}\bigg|^{\frac{1}{n}}=\bigg|\frac{x}{2}\bigg|\frac{1}{n^{\frac{1}{n}}} \to \bigg|\frac{x}{2}\bigg|$$
For which $x$ is this limit less than 1? Consider the boundary cases separately.
For the second part, apply the Weierstrass M-test. Specifically, if $-2<a<b<2$, then then define $t=\max \{ |a| , |b| \}<2$. Then for any $x \in [a,b]$ we have that $|x|<t$ so that $$\bigg|\frac{x^n}{n2^n}\bigg| < \bigg(\frac{t}{2}\bigg)^n$$
Since $t<2$, the series $\sum_n (\frac{t}{2})^n$ converges absolutely, so the M-test tells you that the sequence of partial sums from the original sequence of functions converges uniformly on the given interval $[a,b]$