Chapter 8.1, Problem 40E

To determine

To Find : If the sequence is convergent or divergent using a graph of sequence and then find the limit of the sequence.

Answer to Problem 40E

The sequence is convergent and it’s limit is zero.

Explanation of Solution

Given information : The n^thterm of the sequence is given by, $a_n=\frac{1.3.5.....(2n-1)}{{(2n)}^n}$ .….(1)

By substituting the value of n in equation (1), gives the first five terms as

  $a_1=\frac{1}{{(2\times 1)}^1}=\frac{1}{2}$

  $a_2=\frac{1\times 3}{{(2\times 2)}^2}=\frac{3}{16}$

  $a_3=\frac{1\times 3\times 5}{{(2\times 3)}^3}=\frac{5}{72}$

  $a_4=\frac{1\times 3\times 5\times 7}{{(2\times 4)}^4}=\frac{105}{4096}$

  $a_5=\frac{1\times 3\times 5\times 7\times 9}{{(2\times 5)}^5}=\frac{189}{20,000}$

Table formed with the obtained values is shown below.

The graph of sequence is obtained as shown below, with n in x - axis and a_n in y - axis .

n	an
1	$\frac{1}{2}$
2	$\frac{3}{16}$
3	$\frac{5}{72}$
4	$\frac{105}{4096}$
5	$\frac{189}{20000}$

  

Above graph shows that the sequence is convergent sequence. And its limit is 0.

Proof of limit of a_n ,

  $a_n=\frac{1.3.5.....(2n-1)}{{(2n)}^n}=\frac{\frac{(2n)!}{2^n\times n!}}{{(2n)}^n}$

  $=\frac{(2n)!}{2^n\times n!\times {(2n)}^n}$

Therefore,

  $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{(2n)!}{2^n\times n!\times {(2n)}^n}$

  $=\underset{n\to \infty }{\lim }\frac{1}{2^n\times n!}\times \underset{n\to \infty }{\lim }\frac{(2n)!}{{(2n)}^n}$

  $=0\times 0=0$

Thus,

The obtained sequence is convergent and it’s limit is zero.