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, we can use various tests such as the Ratio Test and the Alternating Series Test (AST). Let's go through each of the given series one by one:
1. The series from n=0 to infinity of ((-1)^n)/n!
To determine if this series is absolutely convergent, we need to find the limit of the absolute value of the ratio of successive terms as n approaches infinity. In this case, the ratio is |((-1)^(n+1)/((n+1)!))/( (-1)^n/n!)| = (n!)/(n+1)!. Simplifying this further, we get |1/(n+1)|, which converges to 0 as n approaches infinity. Since the limit is less than 1, the series is absolutely convergent. Moreover, since an absolutely convergent series is also convergent, we can conclude that it is convergent too.
2. The series from n=0 to infinity of (-1)^n/(sqrt(n^2+n+1))
For this series, let's apply the Ratio Test. Taking the absolute value of the ratio of successive terms, we get |((-1)^(n+1)/(sqrt((n+1)^2+(n+1)+1)))/((-1)^n/(sqrt(n^2+n+1)))|. Simplifying this further, we get |(-1)/(sqrt((n+1)^2+(n+1)+1)/sqrt(n^2+n+1))| = 1/sqrt((n+1)^2+(n+1)+1). As n approaches infinity, the denominator grows without bound, so the limit of the ratio is 0. Since the limit is less than 1, the Ratio Test is inconclusive. However, we can use the Alternating Series Test (AST) in this case. The series is alternating since each term is multiplied by (-1)^n. Also, the absolute value of each term decreases as n increases. Therefore, the series is convergent according to the AST. However, it is not absolutely convergent.
3. The series from n=1 to infinity of (-1)^(n+1)/n^4
To determine the convergence of this series, we can use the Ratio Test again. Taking the absolute value of the ratio of successive terms, we get |((-1)^(n+2)/(n+1)^4)/((-1)^(n+1)/n^4)| = (n^4)/(n+1)^4. As n approaches infinity, the ratio converges to 1, which is not less than 1. Therefore, the series is not absolutely convergent. However, we can use an alternate method to check convergence. Notice that the series is an alternating series similar to the previous example. Moreover, the series term approaches 0 as n approaches infinity. Therefore, we can conclude that the series is convergent according to the AST.
In summary:
1. The series is absolutely convergent and convergent.
2. The series is convergent but not absolutely convergent.
3. The series is convergent but not absolutely convergent.