Chapter 8.1, Problem 3E

To determine

To Find: The first six term of the sequence $a_n=\frac{n}{2n+1}$ and define if the sequence appear to have a limit.

Answer to Problem 3E

The first six term of the sequence is $\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{5}{11},\frac{6}{13}$ and the sequence don’t appear to have a limit as the values are not converging.

Explanation of Solution

Calculation:

Consider the given expression is,

  $a_n=\frac{n}{2n+1}$

Then, the first term is,

  $\begin{array}{c}{a}_1=\frac{1}{2\left(1\right)+1}\\ =\frac{1}{3}\end{array}$

Second term is,

  $\begin{array}{c}{a}_2=\frac{2}{2\left(2\right)+1}\\ =\frac{2}{5}\end{array}$

Third term is,

  $\begin{array}{c}{a}_3=\frac{3}{2\left(3\right)+1}\\ =\frac{3}{7}\end{array}$

Fourth term is,

  $\begin{array}{c}{a}_4=\frac{4}{2\left(4\right)+1}\\ =\frac{4}{9}\end{array}$

Fifth term is,

  $\begin{array}{c}{a}_5=\frac{5}{2\left(5\right)+1}\\ =\frac{5}{11}\end{array}$

Sixth term is,

  $\begin{array}{c}{a}_6=\frac{6}{2\left(6\right)+1}\\ =\frac{6}{13}\end{array}$

Thus, the first six term of the sequence is $\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{5}{11},\frac{6}{13}$ and the sequence don’t appear to have a limit as the values are not converging.