test the series for convergence or divergence.
the sum from n=1 to infinity of ((-1)^n*e^n)/(n^3)
I said it converges because the derivative of (1/n^3) is decreasing
To test the convergence or divergence of the series, you can use the Ratio Test or the Alternating Series Test. Let's use the Ratio Test in this case.
The Ratio Test states that for a series ∑aₙ to converge, the following condition must hold:
lim(n→∞) |aₙ₊₁ / aₙ| < 1.
For the given series: ∑[(-1)^n * e^n / (n^3)], apply the Ratio Test to examine the convergence.
Let's calculate the limit of the ratio |aₙ₊₁ / aₙ| as n approaches infinity:
aₙ = (-1)^n * e^n / (n^3)
aₙ₊₁ = (-1)^(n+1) * e^(n+1) / ((n+1)^3)
Calculating the ratio: |aₙ₊₁ / aₙ|
= [(-1)^(n+1) * e^(n+1) / ((n+1)^3)] / [(-1)^n * e^n / (n^3)]
= [(-1)^(n+1) * e^(n+1) * n^3] / [(-1)^n * e^n * (n+1)^3]
= [(-1)^(n+1) * e^(n+1) * n^3] / [(-1)^n * e^n * (n+1)(n+1)(n+1)]
= [(-1)^(n+1) * e^(n+1) * n^3] / [(-1)^n * e^n * (n+1)^3]
To simplify, we can cancel out the common factors of (-1)^(n+1) and e^(n+1) / e^n:
= [n^3] / [(n+1)^3]
= [n^3] / [n^3 + 3n^2 + 3n + 1]
= 1 / [1 + 3/n + 3/n^2 + 1/n^3]
Now, as n approaches infinity, the expression approaches 1 / 1 = 1.
Since the limit of |aₙ₊₁ / aₙ| is equal to 1, the Ratio Test is inconclusive.
Inconclusive results mean that the test doesn't provide a definitive answer regarding convergence or divergence. Therefore, we need to try a different test.
Given that the series ∑[(-1)^n * e^n / (n^3)] contains alternating terms, we can apply the Alternating Series Test to determine convergence or divergence.
The Alternating Series Test states that if a series ∑(-1)^n * bₙ satisfies the following two conditions:
1. bₙ ≥ 0 for all n
2. bₙ is a decreasing sequence (bₙ₊₁ ≤ bₙ) for all n
Then, the series ∑(-1)^n * bₙ is convergent.
In our case, bₙ = e^n / (n^3).
1. bₙ ≥ 0: This condition is satisfied because e^n and (n^3) are both positive terms for all n.
2. bₙ is a decreasing sequence: To check this condition, let's examine the behavior of bₙ₊₁ ≤ bₙ.
bₙ₊₁ = e^(n+1) / ((n+1)^3)
bₙ = e^n / (n^3)
To show bₙ₊₁ ≤ bₙ, we need to prove that e^(n+1) / ((n+1)^3) ≤ e^n / (n^3).
Dividing both sides by e^n and multiplying both sides by (n^3) gives:
e / (n+1)^3 ≤ 1/n^3
Since the derivative of 1/n^3 is decreasing, we can conclude that 1/n^3 is a decreasing sequence.
Therefore, the condition for the Alternating Series Test is satisfied, and the series ∑[(-1)^n * e^n / (n^3)] is convergent based on the Alternating Series Test.
In summary, the given series converges.