Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 42

Answer

Diverges

Work Step by Step

Given $$ \sum_{n=1}^{\infty} \frac{2^{n^{2}}}{n !}$$ We use the comparison test with: $$\frac{2^{n^{2}}}{n !}\geq \frac{\left(2^{n}\right)^n}{n ^n} $$ Then, by using the $ n$th Root Test, we get: \begin{align*} \rho&= \lim_{n\to \infty} \sqrt[n]{ a_n}\\ &= \lim_{n\to \infty} \sqrt[n]{\frac{\left(2^{n}\right)^n}{n ^n} }\\ &= \lim_{n\to \infty} \frac{2^n}{n}\\ &=\lim_{n\to \infty} \frac{\ln 22^n}{1}\\ &=\infty \end{align*} Thus, by the comparison test, the series $ \sum_{n=1}^{\infty} \frac{2^{n^{2}}}{n !} $ also diverges.
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