Chapter 12.4, Problem 45E

To determine

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

Answer to Problem 45E

The first five terms of given sequence are $1,\frac{3}{5},\frac{2}{5},\frac{5}{17},\frac{3}{13}$ and its limit is $0$ .

Explanation of Solution

Given information:

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

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

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

  $\begin{array}{l}forn=3,{\mathrm{a}}_3=\frac{3+1}{3^2+1}\\ =\frac{4}{9+1}=\frac{4}{10}\\ \Rightarrow {\mathrm{a}}_3=\frac{2}{5}\end{array}$

  $\begin{array}{l}forn=4,{\mathrm{a}}_4=\frac{4+1}{4^2+1}\\ =\frac{5}{16+1}\\ \Rightarrow {\mathrm{a}}_4=\frac{5}{17}\\ forn=5,{\mathrm{a}}_5=\frac{5+1}{5^2+1}\\ =\frac{6}{25+1}\\ \Rightarrow {\mathrm{a}}_5=\frac{6}{26}=\frac{3}{13}\end{array}$

Thus, first five terms are $1,\frac{3}{5},\frac{2}{5},\frac{5}{17},\frac{3}{13}$

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

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

Hence, the first five terms of given sequence are $1,\frac{3}{5},\frac{2}{5},\frac{5}{17},\frac{3}{13}$ and its limit is $0$ .