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∞( An )_____________________Note: In order to get credit for this problem all answers must be correct.please show all steps. I must be makeing asilly mistake. thanks

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Answer:

The sequence is convergent and the serie is divergent.

Step-by-step explanation:

Sequence:

We calculate the limit when n tends to infinity to see if it is convergent or divergent:

[tex]\lim_{n \to \infty} \frac{6n}{-4n+9} = \frac{6}{-4} = \frac{-3}{2} \neq 0[/tex] then, the sequence is convergent.

Serie:

[tex]\sum_{n=1}^{\infty} (\frac{6n}{-4n+9})[/tex]. To know if it converges we are going to use the limit test:

[tex]\lim_{n \to \infty} a_n[/tex]. If this limit is non equal to 0 the serie diverges.

[tex]\lim_{n \to \infty} \frac{6n}{-4n+9} = \frac{6}{-4} = \frac{-3}{2} \neq 0[/tex], then the serie diverges.

Since the limit of the function gives a finite value, hence the seqence converges

Convergence and Divergence

Give the sequence of function

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

In order to check whether the sequence diverges or converges, we will take the limit of the sequence as n tends to infinity as shown below:

║[tex]\lim_{n \to \infty} \frac{6n}{-4n+9}[/tex]

Divide through by n to have:

[tex]\lim_{n \to \infty} \frac{6n/n}{-4n/n+9/n}\\\lim_{n \to \infty} \frac{6}{-4+9/n}[/tex]

Substitute n as infinity to have;

[tex]\lim_{n \to \infty} \frac{6}{-4+0} = -6/4\\\lim_{n \to \infty} \frac{6n}{-4n+9}=-3/2[/tex]

Since the limit of the function gives a finite value, hence the seqence converges

Learn more on convergence here:

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