Chapter 12.4, Problem 47E

To determine

To find the first five terms of the sequence ${\mathrm{a}}_n=\frac{n}{2n+1}$ and its limit.

Answer to Problem 47E

The first five terms of given sequence are $\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{5}{11}$ and its limit is $\frac{1}{2}$ .

Explanation of Solution

Given information:

Given ${\mathrm{a}}_n=\frac{n}{2n+1}$

The first five terms of the sequence ${\mathrm{a}}_n=\frac{n}{2n+1}$ are obtained by substituting $n=1,2,3,4,5$ in general form of ${\mathrm{a}}_n$ .

  $\begin{array}{l}Thus,forn=1,{\mathrm{a}}_1=\frac{1}{2\left(1\right)+1}\\ \Rightarrow {\mathrm{a}}_1=\frac{1}{3}\\ forn=2,{\mathrm{a}}_2=\frac{2}{2\left(2\right)+1}\\ =\frac{2}{4+1}\\ \Rightarrow {\mathrm{a}}_2=\frac{2}{5}\end{array}$

  $\begin{array}{l}forn=3,{\mathrm{a}}_3=\frac{3}{2\left(3\right)+1}\\ =\frac{3}{6+1}\\ \Rightarrow {\mathrm{a}}_3=\frac{3}{7}\end{array}$

  $\begin{array}{l}forn=4,{\mathrm{a}}_4=\frac{4}{2\left(4\right)+1}\\ =\frac{4}{8+1}\\ \Rightarrow {\mathrm{a}}_4=\frac{4}{9}\\ forn=5,{\mathrm{a}}_5=\frac{5}{2\left(5\right)+1}\\ =\frac{5}{10+1}\\ \Rightarrow {\mathrm{a}}_5=\frac{5}{11}\end{array}$

Thus, the first five terms are $\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{5}{11}$

The limit of the sequence is limit of ${\mathrm{a}}_n=\frac{n}{2n+1}$ as $n\to \infty$

  $\begin{array}{l}i.e.\underset{n\to \infty }{\lim }{\mathrm{a}}_n=\underset{n\to \infty }{\lim }\frac{n}{2n+1}\\ =\underset{n\to \infty }{\lim }\frac{n}{2n\left(1+\frac{1}{2n}\right)}\\ =\underset{n\to \infty }{\lim }\frac{1}{2}\left(\because \underset{n\to \infty }{\lim }\frac{1}{n}=0\right)\\ \therefore \underset{n\to \infty }{\lim }{\mathrm{a}}_n=\frac{1}{2}\end{array}$

Hence, the first five terms of given sequence are $\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{5}{11}$ and its limit is $\frac{1}{2}$ .