## Calculus (3rd Edition)

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.