Question

How do you do this question?

Answer 1

Limit = 2

Explanation:

`\sum _(n=1)^(\infty )(2^n)/(3^n)\\= \sum _(n=1)^(\infty \:)\left((2)/(3)\right)^n\\\\= \sum _(n=0)^(\infty \:)\left((2)/(3)\right)^n-\left((2)/(3)\right)^0`

`\sum _(n=0)^(\infty \:)\left((2)/(3)\right)^n=3,\\\\= 3-\left((2)/(3)\right)^0\\= 2`

In this case it does converge, as 2/3 is between 1 and 0

Answer 2

0

Explanation:

2ⁿ / 3ⁿ = (⅔)ⁿ

⅔ is less than 1, so as n approaches infinity, the sequence approaches 0.