Chapter 8.1, Problem 18E

To determine

To find: To determine whether the sequence converges or diverges and find limit of converges.

Answer to Problem 18E

The series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ is converges and its value is ${\displaystyle \sum _{n=0}^{\infty }{a}_n=\frac{1}{3}}$

Explanation of Solution

Given information: $a_n=\sqrt{\frac{n+1}{a_n+1}}$

Solution:

To determine converges or diverges we need to check the limit of the sequence of the partial sums.

  $\underset{n\to \infty }{\lim }{s}_n=\sqrt{\frac{n+1}{a_n+1}}$

Divide numerator and denominator by $\text{'}n\text{'}$ , we will get

  $\begin{array}{l}\Rightarrow \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\sqrt{\frac{\frac{n+1}{n}}{\frac{an+1}{n}}}\\ \Rightarrow \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\sqrt{\frac{1+\frac{1}{n}}{a+\frac{1}{n}}}\left[\underset{n\to \infty }{\lim }\frac{1}{n}=0\right]\\ \Rightarrow \underset{n\to \infty }{\lim }{s}_n=\sqrt{\frac{1}{9}}\\ \Rightarrow \underset{n\to \infty }{\lim }{s}_n=\frac{1}{3}\end{array}$

  $\therefore$ Limit exists and is finite.

Hence, The series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ is converges and its value is ${\displaystyle \sum _{n=0}^{\infty }{a}_n=\frac{1}{3}}$