proof continuity of $f(x)=\sum_{k=1}^{\infty} \frac{|x|^\frac{1}{k}}{2^k}$

Question

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.

Answer 1

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.

Answer 2

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.