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Math Help - Alternating series

  1. #1
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    Alternating series

    Two problems:

    1. ((infinity) summation (n=1)) (((-1)^n)*(n+2))/(n*(n+1))

    Since the top has only one n and the denominator has 2, the fraction will decrease, the limit is 0, and is positive. This converges.

    I then used the ratio test to see if this function converges absolutely...

    I replaced the n(s) with n+1 and got:

    ((n+3)*n)/(n+2)^2... I then took the limit when n approaches infinity and got
    n^2/n^2 = 1... since 0<1<oo, this converges absolutely. Did I do this right?

    2. ((infinity) summation (n=1)) (((-1)^k) * n^3)/ (4^n)

    The issue with this problem is that I never dealt with a problem with two variable as with this one. If anybody can help me, I will be grateful.
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  2. #2
    Newbie Riyzar's Avatar
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    I believe number one converges conditionally. I used the limit comparison test with \frac {1}{n} and got 1. Because \frac {1}{n} diverges \frac {n+2}{n(n+1)} will too.

    For number 2, if a series converges absolutely it will converge no matter what. So if you ignore the alternating portion does \sum \frac {n^3} {4^n} converge? If yes you know that the inclusion of the alternation portion will make no difference. (assuming k is a constant number)
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  3. #3
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    I redid both and for #1 I used the limit comparison and got 1 as well and had the same conclusions.

    for #2, It did converge by alt. series test. for the second part, I used the ratio test to see if it converges absolutely. In the end, I got 6n/24n = 1/4 <1, so it converges absolutely. Did I do this one wrong?

    This is like a side question... Is the alt. series about using the Alt series test, then if it converges (via the guidelines), use another test (ratio test, limit comparison test, etc.) to see if the absolute value of it is absolute or conditional?
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  4. #4
    Newbie Riyzar's Avatar
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    Correct series #2 converges absolutely.

    For your second question yes. If however, the alternating series test fails then you know that the series diverges absolutely and do not have to perform another test.
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  5. #5
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    Thank you Riyzar for your help!! I understand this a heck of a lot more!
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