## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 556: 28

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.

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