Chapter 8.4, Problem 35E

To determine

To prove: ${\displaystyle \sum a_n}$ is convergent or not.

Answer to Problem 35E

  ${\displaystyle \sum a_n}$ is divergent(by D’Alembert’s ratio test).

Explanation of Solution

Given information: $a_1=2,a_{n+1}=\frac{5n+1}{4n+3}{a}_n$

Now, by D’Alembert’s ratio test

  $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{\frac{5n+1}{4n+3}{a}_n}{a_n}\\ =\underset{n\to \infty }{\lim }\frac{5n+1}{4n+3}\\ \underset{n\to \infty }{\lim }\frac{5+\frac{1}{n}}{4+\frac{3}{n}}=\frac{5}{4}>1[\because \underset{n\to \infty }{\lim }\frac{1}{n}=0]\end{array}$

Since, by ratio test $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}>1$

Hence, ${\displaystyle \sum a_n}$ is divergent(by D’Alembert’s ratio test).