Let $E$ be the set of all real number such that $\sum_{n = 1}^{\infty} \frac{x^{2n}}{n^24^n}$ converges absolutely. For each $x \in E$, let $f(x) = \sum_{n = 1}^{\infty} \frac{x^{2n}}{n^24^n}$. Show that $f(x)$ is continuous on E.
I've shown that $E = [-2, 2]$, and I've proceeded to show continuity as follows: Fix some $\epsilon > 0$. Now consider some $x, y \in E, |x|, |y| < 2$. We have:
$\begin{aligned} |f(x) - f(y)| &\leq \sum_{n = 1}^{\infty} \frac{\mid x^{2n} - y^{2n} \mid}{n^24^n} \\ &= \sum_{n = 1}^{\infty} \frac{\mid (x - y)(x^{2n - 1} + x^{2n -2}y + \cdots + y^{2n - 1})\mid}{n^24^n} \\ &\leq \sum_{n = 1}^{\infty} \frac{\mid (x - y) \cdot 2n \cdot 2^{2n - 1}\mid}{n^24^n} \\ &= \,\mid x - y \mid \sum_{n = 1}^{\infty} \frac{1}{n} \end{aligned}$
I'm trying to bound the right hand side of this inequality to get some $\delta(\epsilon)$ and claim continuity, but clearly the harmonic series doesn't converge and I've gone wrong somewhere. Could someone kindly help me complete/fix the proof?
Given $\epsilon >0,$ choose $n_0\in \Bbb N$ such that $\sum_{n=1+n_0}^{\infty}2/n^2<\epsilon/2.$
The polynomial $P_{n_0}(x)=\sum_{n=1}^{n_0}x^{2n}/(n^24^n)$ is continuous on $\Bbb R$ so it is uniformly continuous on $[-2,2].$ So take $\delta>0$ such that $$\forall x,y\in [-2,2]\, (\,|x-y|<\delta\implies |P_{n_0}(x)-P_{n_0}(y)|<\epsilon/2\,).$$
Now whenever $x,y\in [-2,2]$ with $|x-y|<\delta,$ we have $$|f(x)-f(y)=|P_{n_0}(x)-P_{n_0}(y)+\sum_{n=1+n_0}^{\infty}(x^{2n}-y^{2n})/(n^24^n)\,|\le $$ $$\le |P_{n_0}(x)-P_{n_0}(y)|+\sum_{n=1+n_0}^{\infty}(|x|^{2n}+|y|^{2n})/(n^24^n)<$$ $$<\epsilon /2+\sum_{n=1+n_0}^{\infty}(2\cdot 4^n)/(n^24^n)<\epsilon.$$