Question

Given:

Answer 1

Answer:

a.Converge

b.Diverge

Step-by-step explanation:

We are given that

$A_n=\frac{6n}{-4n+9}$

$\lim_{n\rightarrow \infty}A_n=\lim_{n\rightarrow \infty}\frac{6n}{-4n+9}=\frac{6n}{n(-4+\frac{9}{n})}$

$\lim_{n\rightarrow \infty}\frac{6}{-4+\frac{9}{n}}=\frac{6}{-4}=\frac{-3}{2}$

Because $\frac{9}{\infty}=0$

Hence, sequence An converges to $\frac{-3}{2}$ when n tends to infinity or - infinity.

$\sum_{n=1}^{\infty}An=\sum_{n=1}^{\infty}\frac{6n}{-4n+9}$=diverges

Necessary condition for a series to converge :

$an\rightarrow 0$ when n tends to infinity .

But $A_n$ tends to $\frac{-3}{2}$when n tends to infinity.

Therefore, given series is divergent.