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

is this true?

To determine the convergence or divergence of a series, we can use various tests. One commonly used test is the Alternating Series Test, which applies to alternating series of the form ∑((-1)^(n-1) * a_n). In this case, our series fits the form of an alternating series, so we can utilize this test.

The Alternating Series Test states that if the terms in an alternating series decrease in absolute value and approach zero, then the series converges. In other words, if the following conditions are met:
1) |a_n+1| ≤ |a_n| for all n.
2) lim(n→∞) a_n = 0.

In our series, we have a_n = ((-1)^n * e^n) / (n^3). So let's analyze its conditions:

1) To check if the terms decrease in absolute value, we can consider the ratio of terms: |a_n+1 / a_n|. Simplifying this ratio gives us the absolute value of the ratio |((n^3) / ((n+1)^3) * e^(n+1) / e^n)| = |(n/(n+1))^3 * e|. As the exponent e is a positive constant, we only need to focus on the first part, (n/(n+1))^3. If this term is less than or equal to 1 for all n, then our series satisfies the first condition.

To prove this, we can consider the function f(x) = (x/(x+1))^3 for x ≥ 1. Taking the derivative of f(x) and simplifying, we find that f'(x) = (-3x^2) / (x+1)^4, which is negative for x ≥ 1. Therefore, f(x) is a decreasing function for x ≥ 1. Since our series involves positive terms (e.g., |a_n| = a_n), this implies that |a_n+1| ≤ |a_n| holds for all n.

2) To check if the terms approach zero, we need to examine the limit as n approaches infinity: lim(n→∞) ((-1)^n * e^n) / (n^3). This limit evaluates to 0, as the exponential term e^n grows much faster than the polynomial term n^3. Therefore, our series satisfies the second condition as well.

Thus, we can conclude that the series ∑((-1)^n * e^n) / (n^3) converges based on the Alternating Series Test.