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Infinite sum with k^(1/k)

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that
    [tex]\sum_{k=0}^{\infty} \sqrt[k]k-1 [/tex]

    converges.

    2. Relevant equations

    Ratio, radix theorems, comparison with other sums...


    3. The attempt at a solution

    No idea whatsoever.
    Where does one begin in this case ? With other cases I'm quite confident.
     
  2. jcsd
  3. Nov 11, 2012 #2
    Hint:
    Let the general term be denoted by:
    [tex]
    a_k = \sqrt[k]{k} - 1
    [/tex]
    Do you know how to prove that [itex]\lim_{k \rightarrow \infty}{a_k} = 0[/itex]?
    Then, we have:
    [tex]
    k = (1 + a_k)^{k}
    [/tex]
    Taking the same equality for [itex]k + 1[/itex], and subtracting this one, you ought to get:
    [tex]
    1 = (1 + a_{k + 1})^{k + 1} - (1 + a_k)^{k}
    [/tex]
    Solve this equation for [itex]a_{k + 1}[/itex]. What do you get?

    Because as [itex]k \rightarrow \infty[/itex], [itex]a_k[/itex] is an infinitesimal quantity, you may expand your expression for [itex]a_{k + 1}[/itex] in powers of [itex]a_{k}[/itex] up to the first non-vanishing order. What do you get?

    The solution for this asymptotic recursion relation would give you a comparison general term [itex]b_n[/itex], and the series [itex]\sum_{n = 1}^{\infty}{b_n}[/itex] is pretty easy to test for convergence. Then, you may use the [STRIKE]Ratio comparison test[/STRIKE].

    EDIT:
    Use the Asymptotic comparison test mentioned here instead.
     
    Last edited: Nov 11, 2012
  4. Nov 11, 2012 #3
    I don't quite follow you.

    I sort of need simpler methods, thanks anyway.
     
  5. Nov 11, 2012 #4
    Yep, asymptotic comparison seems a good way.
     
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