How do you do this question?

Answers:

[tex]\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = 6[/tex]

Based on the limit comparison test, the series is convergent

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Explanation:

As n gets really large, the terms [tex]3^n[/tex] and [tex]3^n-2[/tex] are effectively the same. Taking the limit to infinity of the ratio leads to 1.

In other words,

[tex]\displaystyle \lim_{n \to \infty} \frac{3^n}{3^n-2} = \lim_{n \to \infty} \frac{3^n}{3^n} = 1[/tex]

That -2 doesn't really play a role when n gets very large.

This result is then multiplied by the 6 in the [tex]a_n[/tex] sequence, getting 6 as the final limit we want.

So,

[tex]L = \displaystyle \lim_{n \to \infty} \left(a_n \div b_n\right)\\\\\\L = \displaystyle \lim_{n \to \infty} \left(\frac{6}{3^n-2} \div \frac{1}{3^n}\right)\\\\\\L = \displaystyle \lim_{n \to \infty} \left(\frac{6}{3^n-2} \times \frac{3^n}{1}\right)\\\\\\L = \displaystyle \lim_{n \to \infty} \left(\frac{6*3^n}{3^n-2}\right)\\\\\\[/tex]

[tex]L = \displaystyle 6*\lim_{n \to \infty} \left(\frac{3^n}{3^n-2}\right)\\\\\\L = \displaystyle 6*\lim_{n \to \infty} \left(\frac{3^n}{3^n}\right)\\\\\\L = \displaystyle 6*\lim_{n \to \infty} (1)\\\\\\L = \displaystyle 6*1\\\\\\L = \displaystyle 6\\\\\\[/tex]

Since the limiting value is positive and not infinity, this means that both series [tex]\sum a_n[/tex] and [tex]\sum b_n[/tex] converge or they both diverge.

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If we plug n = 2 into the [tex]b_n[/tex] sequence, we get

[tex]b_n = \frac{1}{3^n}\\\\b_2 = \frac{1}{3^2}\\\\b_2 = \frac{1}{9}[/tex]

The first term is a = 1/9

If we plugged n = 3, we would get [tex]b_3 = \frac{1}{27}[/tex]

This shows that the common ratio is r = 1/3

Because -1 < r < 1 is true, we know that the infinite geometric series converges. Therefore, [tex]\sum b_n[/tex] converges. We don't need to find the converging value.

Going back to the limit comparison test, we stated that both [tex]\sum a_n[/tex] and [tex]\sum b_n[/tex] either diverge together or converge together.

We've shown that the b series converges, so the 'a' series must converge as well.

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