For what values of p is this series convergent?
(summation from n = 1 to infinity) of ((-1)^(n-1))/(n^(p + 2))
A. p >= -2
B. p =/= -2
C. p > -2
D. for all p
E. p > 0
You have to use the Alternating Series Test.
I've already tried E, and it was wrong. I have one more attempt. I think it's C, but I want to make sure it's correct before I pick it. Thanks!
The p+2 just confuses things. Let q = p+2. Then you have the series
1/1 - 1/2^q + 1/3^q - 1/4^q
The alternating series test says this converges for q>0
That is p > -2
Thanks, Steve!
To determine the convergence of a series, we can use the Alternating Series Test.
According to the Alternating Series Test, for a series of the form Σ((-1)^(n-1)*a(n)), where a(n) is a positive, decreasing, and converging sequence, the series converges if and only if the limit of a(n) as n approaches infinity is 0.
In your case, the series is Σ((-1)^(n-1))/(n^(p+2)), so we need to check if the limit of ((-1)^(n-1))/(n^(p+2)) as n approaches infinity is 0.
Let's analyze the terms of the series:
a(n) = ((-1)^(n-1))/(n^(p+2))
When n is odd, the term becomes ((-1)^0)/(n^(p+2)) = 1/(n^(p+2)).
When n is even, the term becomes ((-1)^1)/(n^(p+2)) = -1/(n^(p+2)).
Now, let's consider two cases based on the value of p:
Case 1: p > -2
In this case, as n approaches infinity, the value of n^(p+2) increases without bound. For both even and odd terms, 1/(n^(p+2)) approaches 0. Therefore, the series converges for p > -2.
Case 2: p <= -2
In this case, as n approaches infinity, the value of n^(p+2) approaches infinity. For both even and odd terms, 1/(n^(p+2)) approaches 0. However, the alternating signs of the terms do not change the fact that the absolute value of the terms does not converge to 0. Therefore, the series diverges for p <= -2.
Based on these cases, we can conclude that the series converges for p > -2. Therefore, the correct answer is C: p > -2.
I hope this helps!