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
Hi:
The series is alternating of the form(a_n)*(-1)^n, where a_n = e^n/n^3. Because limit (n->inf)[a_n] not= 0 (a_n->inf as n->inf}, the series diverges.
Regards,
Rich B.
To determine whether the series converges or diverges, we need to use a convergence test. In this case, we can use the Alternating Series Test, as the series includes the alternating sign (-1)^n.
The Alternating Series Test states that if an alternating series satisfies two conditions:
1. The terms alternate in sign.
2. The absolute value of each term decreases as n increases.
Let's analyze each condition for the given series:
1. The terms alternate in sign:
The series has the alternating sign (-1)^n, which means the sign of each term changes as n changes.
2. The absolute value of each term decreases as n increases:
To check this condition, let's examine the absolute value of each term:
|((-1)^n * e^n) / (n^3)| = (e^n / (n^3))
For convergence, it suffices to show that the sequence (e^n / (n^3)) is decreasing.
To determine the trend of this sequence, we can consider the ratio of consecutive terms:
R = [e^(n+1) / ((n+1)^3)] / [e^n / (n^3)]
= (e^(n+1) * (n^3)) / (e^n * (n+1)^3)
= (e * n^3) / ((n+1)^3)
To show that the sequence is decreasing, we need to show that R < 1 for all n.
(e * n^3) / ((n+1)^3) < 1
(e * n^3) < (n+1)^3
Take the cube root of both sides:
e^(1/3) * n < (n+1)
Now, divide both sides by n:
e^(1/3) < (n+1) / n
As n approaches infinity, (n+1)/n approaches 1. Therefore, e^(1/3) < 1.
Since e^(1/3) is a constant less than 1, we have shown that the sequence (e^n / (n^3)) is decreasing, satisfying condition 2 of the Alternating Series Test.
We have established that both conditions of the Alternating Series Test are met. Therefore, the given series converges.