Answer
Converges
Work Step by Step
Given
$$ \sum_{k=1}^{\infty} \frac{1}{2^{k^{2}}}$$
Compare with $\displaystyle\sum_{k=1}^{\infty}\frac{1}{2^{k}} $, which is a convergent series ( geometric with $|r|<1$) and for $k\geq 1$
\begin{align*}
2^{k^{2}} &\geq 2^{k}\\
\frac{1}{2^{k^{2}}}& \leq \frac{1}{2^{k}}
\end{align*}
Then $\displaystyle\sum_{k=1}^{\infty} \frac{1}{2^{k^{2}}}$ also converges.
