determine if the series is absolutely convergent and convergent.
the sum from n=0 to infinity of ((-1)^n*e^n)/(n!)
I used the ratio test and said that it was absolutely convergent and convergent.
is this true?
To determine if the series ((-1)^n*e^n)/(n!) is absolutely convergent and convergent, you correctly used the ratio test. Let's go through the steps to verify if your conclusion is correct.
The ratio test states that if the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term is less than 1, then the series is absolutely convergent.
Let's apply the ratio test to the given series:
First, consider the (n+1)th term:
a(n+1) = ((-1)^(n+1)*e^(n+1))/((n+1)!)
Now, calculate the ratio:
|r| = |((-1)^(n+1)*e^(n+1))/((n+1)!) / ((-1)^n*e^n)/(n!)|
= |(-1)^(n+1)*e^(n+1)*n!/(n+1)! * n!/(e^n*(-1)^n)|
Simplifying the expression:
|r| = |((-1)^(n+1)*e^(n+1)*n!*n!)/(e^n*(n+1)!*(-1)^n)|
= |(-1)^n*e*(n^2)/(n+1)|
Taking the limit of |r| as n approaches infinity:
lim(n->∞) |(-1)^n*e*(n^2)/(n+1)| = ∞
Since the limit is greater than 1, the ratio test indicates that the series is divergent for all values of n. Therefore, the series is not absolutely convergent.
Next, let's consider the convergence of the series without taking the absolute value. Since the series is alternating due to the (-1)^n term, we can use the Alternating Series Test to determine its convergence.
The Alternating Series Test states that if the magnitude (or absolute value) of the terms in an alternating series decrease monotonically to 0, then the series is convergent.
In this series, the terms can be written as (-1)^n*(e^n)/(n!). As n increases, the term approaches 0, because the factorial term in the denominator grows much faster than the exponential term in the numerator.
Therefore, we can conclude that the series ((-1)^n*e^n)/(n!) is convergent.
In summary:
- The series ((-1)^n*e^n)/(n!) is not absolutely convergent based on the ratio test.
- The series ((-1)^n*e^n)/(n!) is convergent based on the Alternating Series Test.
So, your conclusion is correct.