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Homework Help: Calculating where X converges

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine values for x where [tex]\sum \frac{(2^k)(x^k)}{ln(k + 2)}[/tex] converges if k goes from 0 to infinity.

    2. Relevant equations

    Limit test and alternating series test.

    3. The attempt at a solution

    I used the limit comparison test to test out ak+1 and an and I ended up getting [tex](-1/2) \le x \le (1/2)[/tex]

    Then I had to compare the endpoints. I noticed that the numerator is always 1 in the case of x=1/2 or always 1 or -1 (in the case of x = -1/2).

    So by the alternating series test, I concluded that BOTH of the endpoints converges. However, my answer key says that x=1/2 diverges while x=-1/2 converges (I'm correct there).

    FYI, for x = 1/2,

    I said that an+1 < an and that the limit as k -> [tex]\infty[/tex] = 0, so why does it apparently diverge?
  2. jcsd
  3. Mar 11, 2010 #2


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    I'm with you at x=(-1/2), but how did you conclude 1/ln(k+2) converges by the alternating series test? It doesn't even alternate.
  4. Mar 11, 2010 #3
    Wait so I can't use the alternating series test with a non alternating series ... wow I'm going to kick myself in the @%$@# for overlooking this.

    So I assume I just use like a comparison test for 1/ln(k+2) with 1/k (diverges) since ln makes values of k smaller so the frac. gets bigger.
  5. Mar 11, 2010 #4


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  6. Mar 12, 2010 #5
    This is the Ratio Test not the Limit Comparison Test.
    Also, you should write [tex]a_k[/tex] not [tex]a_n[/tex].
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