Is this correct?
determine if absolutely convergent and convergent
1. the series from n=0 to infinity of ((-1)^n)/n!
I said it was abs. conv, and therefore conv
2. the series from n=0 to infinity of
(-1)^n/(the square root of (n^2+n+1))
I said ratio test was inconclusive so not abs. conv. but conv. from the A.S.T.
3. the series from n=1 to infinity of (-1)^(n+1)/n^4
I wasn't sure how to do this one, I know not abs. conv. because the ratio test was one but how do you tell if conv?
To determine if a series is absolutely convergent and convergent, you can use various tests such as the ratio test or the alternating series test (AST).
1. For the series from n=0 to infinity of ((-1)^n)/n!:
- The series is the alternating series, so the ratio test can be applied.
- Apply the ratio test:
lim(n->∞) |((-1)^(n+1)/((n+1)!))/((-1)^n/n!)|
= lim(n->∞) |(-1)^(n+1) * n! / ((n+1)!*(-1)^n)|
= lim(n->∞) |(-1)^(n+1) / (n+1)|
= 0
- Since the limit is less than 1, the series converges absolutely, and therefore it also converges.
2. For the series from n=0 to infinity of (-1)^n/(sqrt(n^2+n+1)):
- The series is the alternating series, so you can use the alternating series test.
- The terms approach zero as n goes to infinity.
- The terms are decreasing because each term alternates in sign and the denominator (sqrt(n^2+n+1)) increases as n increases.
- Therefore, according to the alternating series test, the series converges.
3. For the series from n=1 to infinity of (-1)^(n+1)/n^4:
- To determine if the series is convergent, you can use the convergent series comparison test with a known convergent series.
- Consider the series 1/n^4, which is known to converge (p-series test with p=4 > 1).
- Compare it with the given series: |((-1)^(n+1))/n^4| ≤ 1/n^4
- Since 1/n^4 is known to converge and the absolute value of the terms of the given series is bounded by it, the given series also converges.
In summary:
1. The series from n=0 to infinity of ((-1)^n)/n! is both absolutely convergent and convergent.
2. The series from n=0 to infinity of (-1)^n/(sqrt(n^2+n+1)) is convergent.
3. The series from n=1 to infinity of (-1)^(n+1)/n^4 is convergent.