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?
 A: Note that on the interval $(-2,2)$, 
$$
\frac{x^{2n}}{n^24^n}\leq\frac{4^n}{n^24^n}=\frac{1}{n^2}
$$
which is summable. Then, we can conclude by the dominated convergence theorem
that for $x_0\in(-2,2)$
$$
\lim_{x\rightarrow x_0}f(x)=\lim_{x\rightarrow x_0}\sum_{n=1}^{\infty}\frac{x^{2n}}{n^24^n}=
\sum_{n=1}^{\infty}\lim_{x\rightarrow x_0}\frac{x^{2n}}{n^24^n}=\sum_{n=1}^{\infty}\frac{x_0^{2n}}{n^24^n}=f(x_0)
$$
and $f$ is continuous.
A: You can show that the series converges uniformly, and so the limit function must be continuous (this doesn't hold for pointwise convergence in general).
To show uniform convergence, apply the Weierstrass M-test to the sequence of functions: $f_n:(-2,2)\rightarrow \mathbb{R}$
|$\frac{x^{2n}}{n^{2}4^{n}}$|. We can see that we can find an $M_n$ which allows us to bound the sequence in $(-2,2)$. Note that $|x^{2n}| < 4$ $\forall x$ $\in (-2,2)$. 
by standard comparison, we have that 
|$\frac{x^{2n}}{n^{2}4^{n}}|$ $ < |\frac{4}{n^{2}4^{n}}|$  $ < \frac{4}{n^{2}}$
but we have now that $\sum_{k=1}^{\infty} \frac{4}{n^{2}}$ $\rightarrow \frac{\pi^{2}}{6}$ and so now we have,  by the Weierstrass M-test, that our function defined as a series: 
$f(x)$=$\sum_{k=1}^{\infty}$ $\frac{x^{2n}}{n^{2}4^{n}}$ is uniformly convergent and so our function is continuous. This is also how Weierstrass' famous continuous everywhere, differentiable nowhere function was proven to be continuous. 
Note, you can also argue that the function is analytic, hence $C^{\infty}$, hence continuous. 
A: HINT:
If a function is analytic, it is $C^\infty$ according to Taylor's theorem, and if it is smooth or $C^\infty$...
