• Mar 15th 2009, 12:58 PM
oceanmd
Please look over the following series to determine wheather it is convergent or divergent by specific test. If my answer is correct, please just say the number of the problem and yes, if it is wrong, I would live to have an explanation, but if it's too much work, evn to know that it is wrong will help.
Thank you very, very much.

1) series infinity, n=1 (n^2-6)/(n+4) by N-th term test
My answer: lim of the n-th term does not exist, test failed

2) series infinity, n=1, 5^n/3^n (geometric series)
My answer: r=5/3 greater than 1, geometric series diverges.

3) series infinity, n=1, 1/n^(3/4) (p-series test)
My answer: p=3/4 less than 1, series diverges.

4) series infinity, n=1 (n^2-2n)/3n^3+4n^2) by limit comparison
My answer: both series a(n) and b(n) diverge based on b(n)

5) series infinity, n=1, 2^n/(2n+4)! by ratio test
My answer: lim 1/2n^2 =0 less than 1, converges

6) series infinity, n=1, (n-2)/(3n+4) by alternating test

7) series infinity, n=1, 1/(3n^4+1) by comparison test
• Mar 15th 2009, 01:25 PM
skeeter
Quote:

6) series infinity, n=1, (n-2)/(3n+4) by alternating test ?

diverges because it fails the nth term test.

The rest look ok to me.
• Mar 15th 2009, 01:47 PM
oceanmd
Thank you
skeeter,

Thank you a lot.
• Mar 15th 2009, 02:38 PM
HallsofIvy
Quote:

Originally Posted by oceanmd
Please look over the following series to determine wheather it is convergent or divergent by specific test. If my answer is correct, please just say the number of the problem and yes, if it is wrong, I would live to have an explanation, but if it's too much work, evn to know that it is wrong will help.
Thank you very, very much.

1) series infinity, n=1 (n^2-6)/(n+4) by N-th term test
My answer: lim of the n-th term does not exist, test failed

The test does not "fail". Because it does not have a limit of 0, the sequence diverges.

Quote:

2) series infinity, n=1, 5^n/3^n (geometric series)
My answer: r=5/3 greater than 1, geometric series diverges.

3) series infinity, n=1, 1/n^(3/4) (p-series test)
My answer: p=3/4 less than 1, series diverges.

4) series infinity, n=1 (n^2-2n)/3n^3+4n^2) by limit comparison
My answer: both series a(n) and b(n) diverge based on b(n)
What "a(n)" and "b(n)" are you talking about?

Quote:

5) series infinity, n=1, 2^n/(2n+4)! by ratio test
My answer: lim 1/2n^2 =0 less than 1, converges

6) series infinity, n=1, (n-2)/(3n+4) by alternating test
What "alternating test do you mean? If you mean "If a(n) alternates sign and a(n) decreases to 0, the the series converges" that doesn't apply here. It says nothing about what happens if a(n) is not alternating. As skeeter said, the series diverges because (n-2)/(3n+4) does not go to 0.

Quote:

7) series infinity, n=1, 1/(3n^4+1) by comparison test