#6 If A to the nth term is > 0 for all n and lim n approaches infinity (a to nth term +1)/(a to nth term) = 3, which of the following series converges

Question

#6 If A to the nth term is > 0 for all n and lim n approaches infinity (a to nth term +1)/(a to nth term) = 3, which of the following series converges

1. Series 1 is the lower limit ; ∞ is the upper limit ; function is a to the nth term
2. Series 1 is the lower limit ; ∞ is the upper limit ; function is (a to the nth term)/(n^5)
3. Series 1 is the lower limit ; ∞ is the upper limit ; function is (a to the nth term )/(5^n)
4. Series 1 is the lower limit ; ∞ is the upper limit ; function is (a to the nth term )^2/(5^n)

Answer 1

In this problem, we have a sequence defined by the terms "a to the nth term," and we know that the limit of the ratio of consecutive terms as n approaches infinity is 3. We want to determine which of the following series converges based on different functions of the nth term. We can use the limit comparison test to determine convergence.

The limit comparison test states that if you have two series, Σa_n and Σb_n, and if:

lim (n → ∞) (a_n / b_n) = L, where L is a positive finite number,

then both series Σa_n and Σb_n either both converge or both diverge.

Let's consider each series:

Series 1: Σa_n

Series 2: Σ(a_n / n^5)

Series 3: Σ(a_n / 5^n)

Series 4: Σ(a_n^2 / 5^n)

Given that lim (n → ∞) [(a_n + 1) / a_n] = 3, we will compare each series to Series 1.

1. For Series 1, we have Σa_n.

2. For Series 2, we have Σ(a_n / n^5).

3. For Series 3, we have Σ(a_n / 5^n).

4. For Series 4, we have Σ(a_n^2 / 5^n).

Let's consider Series 2, Series 3, and Series 4 one by one:

For Series 2, as n grows to infinity, (a_n / n^5) goes to zero since a_n is always greater than zero, and n^5 grows much faster. Therefore, it converges.

For Series 3, as n grows to infinity, (a_n / 5^n) also goes to zero because 5^n grows faster than a_n. Therefore, it converges.

For Series 4, as n grows to infinity, (a_n^2 / 5^n) also goes to zero, and since it's squared, it will be even smaller than Series 3. Therefore, it also converges.

So, based on the limit comparison test, all of Series 1, Series 2, Series 3, and Series 4 converge.