Chapter 10.5, Problem 11E

To determine

To find: that the whether the series converges or diverges.

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{{\left(\ln 2\right)}^n}}$

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }{a}_n={\displaystyle \sum _{n=1}^{\infty }\frac{1}{{\left(\ln 2\right)}^n}}}\\ a_n=\frac{1}{{\left(\ln 2\right)}^n}\\ a_{n+1}=\frac{1}{{\left(\ln 2\right)}^{n+1}}\\ \frac{a_{n+1}}{a_n}=\frac{{\left(\ln 2\right)}^n}{{\left(\ln 2\right)}^{n+1}}=\frac{1}{\ln 2}\\ \underset{n\to \infty }{\lim }\left|\frac{a_n+1}{a_n}\right|=\underset{n\to \infty }{\lim }\frac{1}{\ln 2}=\frac{1}{0.693}>1\end{array}$

Therefore, the series is divergent by ratio test

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }{a}_n={\displaystyle \sum _{n=1}^{\infty }\frac{1}{{\left(\ln 3\right)}^n}}}\\ a_n=\frac{1}{{\left(\ln 3\right)}^n}\\ a_{n+1}=\frac{1}{{\left(\ln 3\right)}^{n+1}}\\ \frac{a_{n+1}}{a_n}=\frac{{\left(\ln 3\right)}^n}{{\left(\ln 3\right)}^{n+1}}=\frac{1}{\ln 3}\\ \underset{n\to \infty }{\lim }\left|\frac{a_n+1}{a_n}\right|=\underset{n\to \infty }{\lim }\frac{1}{\ln 3}=\frac{1}{1.732}>1\end{array}$

Therefore, the series is convergent by ratio test