Chapter B, Problem 10AYU

To determine

To find: Determine whether each sequence converges or diverges. If it converges, find its limit.

Answer to Problem 10AYU

Solution:

The given sequence converges. The limit is $\frac{1}{2}$.

Explanation of Solution

Given:

$\left\{c_n\right\} = \left\{\frac{4n^3 + 2n + 7}{8n^3 + n^2 - 2}\right\}$

Calculation:

$\left\{c_n\right\} = \left\{\frac{4n^3 + 2n + 7}{8n^3 + n^2 - 2}\right\}$

$\left\{\frac{4n^3+2n+7}{8n^3+n^2-2}\right\}=\left\{\frac{4n^3(1+\frac{2n}{4n^3}+\frac{7}{4n^3}}{8n^3(1+\frac{n^2}{8n^3}-\frac{2}{8n^3}}\right\}=\frac{1}{2}$

Hence the sequence converges. Therefore we will find the limit.

$\underset{n\to \infty }{\lim }{c}_n=\underset{n\to \infty }{\text{lim}}\left\{\frac{4n^3(1+\frac{2n}{4n^3}+\frac{7}{4n^3}}{8n^3(1+\frac{n^2}{8n^3}-\frac{2}{8n^3}}\right\}=\frac{1}{2}$