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

To determine which of the given series converges, we can use the limit comparison test, which states that if you have two series, ∑a_n and ∑b_n, and:

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

2. If ∑b_n converges, then ∑a_n also converges.

3. If ∑b_n diverges, then ∑a_n also diverges.

In your case, we have the series as described in your question, with a to the nth term:

1. Series 1: ∑(a^n), where a > 0.

2. Series 2: ∑(a^n / n^5).

3. Series 3: ∑(a^n / 5^n).

4. Series 4: ∑(a^(2n) / 5^n).

We are given that:

lim (n → ∞) [(a^(n+1)) / (a^n)] = 3.

Since the limit is a positive finite constant (3), let's compare each series with a known series to see if it converges or diverges.

1. Series 1: ∑(a^n)

This is a geometric series with a common ratio a. It converges if -1 < a < 1 and diverges otherwise. The limit comparison test isn't necessary here. The convergence depends on the value of 'a.'

2. Series 2: ∑(a^n / n^5)

Let's compare it to the p-series ∑(1/n^5). The limit of (a^n / (1/n^5)) as n approaches infinity is a^5.

- If a^5 < 1, then ∑(a^n / n^5) converges.

- If a^5 > 1, then ∑(a^n / n^5) diverges.

3. Series 3: ∑(a^n / 5^n)

This is also a geometric series with a common ratio (a/5). It converges if -1 < (a/5) < 1.

- If -5 < a < 5, then ∑(a^n / 5^n) converges.

- If a ≤ -5 or a ≥ 5, then ∑(a^n / 5^n) diverges.

4. Series 4: ∑(a^(2n) / 5^n)

This is a geometric series with a common ratio (a^2/5). It converges if -1 < (a^2/5) < 1.

- If -√5 < a < √5, then ∑(a^(2n) / 5^n) converges.

- If a ≤ -√5 or a ≥ √5, then ∑(a^(2n) / 5^n) diverges.

In summary, the convergence of these series depends on the value of 'a' and can be determined as follows:

1. Series 1 converges if -1 < a < 1.

2. Series 2 converges if a^5 < 1.

3. Series 3 converges if -5 < a < 5.

4. Series 4 converges if -√5 < a < √5.

The exact convergence conditions for these series depend on the value of 'a'.