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Homework Help: Are these the best tests for convergence of the following series?

  1. Feb 7, 2008 #1
    1. The problem statement, all variables and given/known data
    (a) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}}{n^{1 + \frac{1}{n}}}\Bigg)[/tex]
    (b) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{e^{\frac{1}{n}}}{n^{2}}\Bigg)[/tex]
    (c) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}n!}{n^{n}}\Bigg)[/tex]
    (d) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{1}{n^{1 + \frac{1}{n}}}\Bigg)[/tex]
    (e) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{ln(x)}{n^{\frac{3}{2}}}\Bigg)[/tex]
    (f) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(1 - e^{\frac{-1}{n}}\Bigg)[/tex]

    2. Relevant equations
    The test that we have messed with are:
    Telescoping, Geometric, P-Series, Ratio, Root, Simple Comparison, Limit Comparison, Absolute Convergence, Alternating Series, Dirichlet, Integral, Gauss

    3. The attempt at a solution

    First, I got they all converge:

    (a) (d)
    I worked 'd' first. I did a limit comparison test with 1/n
    [tex]\frac{\frac{1}{n}}{\frac{1}{n^{1 + \frac{1}{n}}}} = n^{n}[/tex]
    That limit is 1. And since its absolute value converges (a) converges.

    (b) Basic Comparion test wtih <= [tex]\frac{e}{n^{2}}[/tex]

    (c) Alternating series test

    (e) Integral test

    (f) Im not sure which test to apply
    Last edited: Feb 7, 2008
  2. jcsd
  3. Feb 7, 2008 #2
    (d) is wrong,a dn so is (a), not sure how to prove if they are convergent (if they are).

    Is (d) divergent?
  4. Feb 7, 2008 #3
    (a) is convergent by alternating series test.

    Did I end up proving (d) divergent or is my proof wrong?

    I'm also quite lost on (f) since I can't integrate that function nicely
  5. Feb 7, 2008 #4


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    Homework Helper

    Too many questions at once! d) is divergent. It's the same as 1/(n*n^(1/n)). For n^(1/n) show that the log of that approaches zero. So n^(1/n) approaches 1. So you can do a comparison with say 1/(2n). For f) expand e^(-1/n) in a power series using e^x=1+x+x^2/2!+etc and keep only the terms that matter.
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