# Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 56

Diverges

#### Work Step by Step

Given $$\sum_{n=1}^{\infty}(0.8)^{-n} \cdot n^{-0.8}=\sum_{n=1}^{\infty}\left(\frac{1}{0.8}\right)^{-n} \cdot \frac{1}{n^{0.8}}=\sum_{n=1}^{\infty} \frac{1.25^{n}}{n^{0.8}}$$ By using the Ratio Test, we get: \begin{aligned} \rho &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \\ &=\lim _{n \rightarrow \infty} \frac{1.25^{n+1}}{(n+1)^{0.8}}\frac{n^{0.8}}{1.25^{n}} \\ &=\lim _{n \rightarrow \infty} (1.25)\left(\frac{n}{n+1}\right)^{0.8} \\ &=1.25>1 \end{aligned} Thus the series diverges.

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