Find the radius of convergence for $\sum_{k=0}^\infty 2^{k} x^{k^2}$

Question

For my homework I have to find the radius of convergence of $\sum_{k=0}^\infty 2^{k} x^{k^2}$, I found :

$$\lim_{k\to\infty} \left|\frac{2^{k+1}x^{\left(k+1 \right)^2}}{2^k x^{k^2}}\right| = \left|\frac{2x^{k^2+2k+1}}{x^{k^2}}\right| = \left| 2x^{2k+1} \right| = 2 \cdot \lim_{k\to\infty} \left| x^{2k+1} \right| = 2 \cdot \left| x^{\infty} \right| $$

Thus the power serie converges only when $x=0$.

Is my answer correct?

Answer 1

No, it is incorrect.

At $x=-1$ and $x=1$, the series is divergent.

For $\forall x\in (-1,1)$, $$2 \lim_{k\to \infty} |x^{2k+1}|=0<1$$ So the convergent radius is $1$.