Prove that a function defined as series is continuous I want to show that the function $f(x) = \sum_{n = 1}^\infty\frac{x^{2n}}{n^24^n}$ is continuous on $(-2, 2)$. I've shown that the series converges on that interval, so how can I show that the function is continuous? Should I proceed straight from the definition? If so, then given $\epsilon > 0$, we have $|f(x_1) - f(x_2)| = \sum_{n = 1}^\infty\frac{x_1^{2n} - x_2^{2n}}{n^24^n}$, which we want to be less than $\epsilon$ whenever $|x_1 - x_2| < \delta$ for some $\delta > 0$. How do I choose $\delta$ to see that this holds?

HINT:
If a function is analytic, it is $C^\infty$ according to Taylor's theorem, and if it is smooth or $C^\infty$...
