Question

What is the value of L = lima, when applying the Root Test to assess 2n2 - 1 convergence of Ź ? What is the conclusion of the test? 5-75-4n2 A. 72;convergent B. 72; divergent C.0;convergent D.2/5;convergent

Answer 1

1. Answer: A. 1/2; convergent.

The series given is:

$\sum_{i=0}^{\infty}a_n=\sum_{i=0}^{\infty}\left(\frac{2n^2-1}{5-4n^2}\right)^n$

So, we have:

$a_n=\left(\frac{2n^2-1}{5-4n^2}\right)^n$

So, we have L as:

$L=\lim_{n\to \infty}\sqrt[n]{|a_n|}=\lim_{n\to \infty}\sqrt[n]{\left|\frac{2n^2-1}{5-4n^2}\right|^n}\\ =\lim_{n\to \infty}\left|\frac{2n^2-1}{5-4n^2}\right| =\lim_{n\to \infty}\left|\frac{2-1/n^2}{5/n^2-4}\right|=\left|\frac{2-0}{0-4}\right|\\ =\left|-\frac{2}{4}\right|=\frac{1}{2}\\ \\ \therefore L=\frac{1}{2}<1$

As L is less than 1, the series is convergent.

2. Answer: B. 0; convergent.

The series given is:

$\sum_{i=1}^{\infty}a_n=\sum_{i=1}^{\infty}\frac{n(-3)^n}{(n+1)!}$

So, we have:

$a_n=\frac{n(-3)^n}{(n+1)!}$

So, we have $\rho$ as:

$\rho=\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to \infty}\left|\frac{(n+1)(-3)^{n+1}}{(n+2)!}\cdot\frac{(n+1)!}{n(-3)^{n}}\right|\\ =\lim_{n\to \infty}\left|(-3)\cdot\frac{n+1}{n}\cdot\frac{\cancel{(n+1)!}}{(n+2)\cancel{(n+1)!}}\right| =\lim_{n\to \infty}\left|(-3)\cdot\left( 1+1/n\right )\cdot\frac{1/n}{1+2/n}\right|\\ =\left|(-3)\cdot (1+0)\cdot\frac{0}{1+0}\right|=|0|=0$

As $\rho$ is equal to 0, the series is convergent.