Given:
An = [6 n/(-4 n + 9)]
For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise.

(a) The sequence {An }._
(b) The series ∑n=1[infinity]( An )

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lublana lublana · Answer 1 · 2019-09-16T09:57:38+02:00

Answer:

a.Converge

b.Diverge

Step-by-step explanation:

We are given that

[tex]A_n=\frac{6n}{-4n+9}[/tex]

[tex]\lim_{n\rightarrow \infty}A_n=\lim_{n\rightarrow \infty}\frac{6n}{-4n+9}=\frac{6n}{n(-4+\frac{9}{n})}[/tex]

[tex]\lim_{n\rightarrow \infty}\frac{6}{-4+\frac{9}{n}}=\frac{6}{-4}=\frac{-3}{2}[/tex]

Because [tex]\frac{9}{\infty}=0[/tex]

Hence, sequence An converges to [tex]\frac{-3}{2}[/tex] when n tends to infinity or - infinity.

[tex]\sum_{n=1}^{\infty}An=\sum_{n=1}^{\infty}\frac{6n}{-4n+9}[/tex]=diverges

Necessary condition for a series to converge :

[tex] an\rightarrow 0[/tex] when n tends to infinity .

But [tex]A_n[/tex] tends to [tex]\frac{-3}{2}[/tex]when n tends to infinity.

Therefore, given series is divergent.