Mathematics, 18.10.2020 14:01, brenmichelle24

How do you do this question?

Answers: 2

Show answers

Answers

answered: bwjester0

Answers:

$\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = 6$

Based on the limit comparison test, the series is convergent

========================================================

Explanation:

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

In other words,

$\displaystyle \lim_{n \to \infty} \frac{3^n}{3^n-2} = \lim_{n \to \infty} \frac{3^n}{3^n} = 1$

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

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

So,

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

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

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

----------------------------

If we plug n = 2 into the b_n sequence, we get

$b_n = \frac{1}{3^n}\\\\b_2 = \frac{1}{3^2}\\\\b_2 = \frac{1}{9}$

The first term is a = 1/9

If we plugged n = 3, we would get $b_3 = \frac{1}{27}$

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, $\sum b_n$ converges. We don't need to find the converging value.

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

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

answered: Guest

However, if the workers do outside of what's normally expected, go ahead and tip.

answered: Guest

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,