proof continuity of $f(x)=\sum_{k=1}^{\infty} \frac{|x|^\frac{1}{k}}{2^k}$ I need to prove that the function
$$f:\mathbb{R} \to \mathbb{R}, f(x)=\sum_{k=1}^{\infty} \frac{|x|^\frac{1}{k}}{2^k}$$
is well defined and continuous.
I think I already got the well defined part by proving convergence with the ratio test. But the continuity gives me a headache, since the series is not uniform convergent (correct me if i am wrong).
P.S.: Sorry for the spelling mistakes, English is not my first language.
 A: Using the comments above, here is a proof.
Inside the interval $[-1,1]$ there is uniform convergence, since you can bound
$$\frac{|x|^{1/k}}{2^k} \le \frac{1}{2^k}$$
This proves that $f$ is continuous in $(-1,1)$.
Now, let's work in a different interval, say $x \in (\alpha , + \infty)$, where $0 < \alpha < 1$ has to be determined yet. Inside this interval, consider
$$\frac{f(x)}{x} = \sum_{k=1}^{\infty} \frac{1}{x^{1-1/k} \cdot 2^k}$$
I want to bound for all $k \ge2$
$$\frac{1}{x^{1-1/k} \cdot 2^k} < \frac{1}{1.5^k} \ \qquad \qquad \qquad \ (*)$$
This gives me the inequality
$$x>  \left( \frac{3}{4} \right)^{k^2/(k-1)}$$
This is satisfied if $$x>  \sup_{k \ge 2} \left( \frac{3}{4} \right)^{k^2/(k-1)} = \left( \frac{3}{4} \right)^{2^2/(2-1)} = \left( \frac{3}{4} \right)^4=0.316...$$
so, if we consider $\alpha = 0.5$ we have that condition $(*)$ is satisfied.
This proves that the series
$$\sum_{k=1}^{\infty} \frac{1}{x^{1-1/k} \cdot 2^k} < \sum_{k=1}^{\infty} \frac{1}{1.5^k}$$
is uniformly convergent to $f(x)/x$ in $( \alpha , + \infty)$.
In particular, $f(x)/x$ is continuous on that interval. Hence
$$f(x) = \frac{f(x)}{x} \cdot x$$
is continuous on that interval.
Since the function is even, and it's continuous at every non-negative real number, it's continuous on the whole real line: this concludes the proof.
A: For $n\in \Bbb N$ let $f_n(x)=\sum_{k=1}^n|x|^{1/k}2^{-k}.$ Show that $f_n\to f$ uniformly 
 ( as $n\to \infty$ ) on the domain $[-m,m],$ for each $m\in \Bbb N,$  using the fact that if $m\in \Bbb N$ and $|x|\le m$ then $|f_n(x)-f(x)|\le \sum_{k=n+1}^{\infty}m2^{-k}.$
Since each $f_n$ is continuous, therefore $f$ is continuous on each $[-m,m],$ and therefore $f:\Bbb R \to \Bbb R$ is continuous.
