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Can someone help me on this series?

  1. Jan 12, 2007 #1
    Hello,
    I want to check convergence for the following series:
    ∑2^(1/n)-1
    so that's the n'th root of two minus one, for n going to positive infinity

    Now all possible formulas i know come out inconclusive,
    so the only formula i can think of is the integer formula, now i thought
    to make things easier to just integrate 2^(1/n) and if this series proves to be convergent then that proves the convergency (as it is always "1" larger than the other one).

    Now as i am seem to be a complete idiot i cannot even find the integer for this one.

    Can someone help me?

    Sorry if my English is not so good.

    Thank you in advance,
    Egwin.
     
  2. jcsd
  3. Jan 12, 2007 #2

    StatusX

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    You have the expansion:

    [tex] (1+x)^p = 1 +p x + O(p^2) [/tex]

    What happens when x=1 and p=1/n is very small?
     
  4. Jan 26, 2007 #3
    thanks

    i didnt really know what you meant,

    but i have the solution now,
    if you're interested, it's in the attachment.

    egwin.
     

    Attached Files:

  5. Feb 2, 2007 #4
    hello status x,

    in your expansion /x/ should be <1
    does it still work for x=1 then?
     
  6. Feb 2, 2007 #5

    StatusX

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    The full expansion is:

    [tex](1+x)^p=1+px+\frac{p(p-1)}{2!} x^2 + \frac{p(p-1)(p-2)}{3!} x^3 + ...[/tex]

    which is valid for all p, all |x|<1. I made a mistake before, the second term is actually still O(p) as p->0. For p very small, we can approximate [itex]p-n \approx -n[/itex], so:

    [tex](1+x)^p \approx 1+px(1+\frac{-1}{2} x + \frac{2\cdot 1}{3!} x^2 + ...)[/tex]

    [tex] =1+px(1-\frac{1}{2} x + \frac{1}{3} x^2 + ...)[/tex]

    with x=1 (or, more accurately, taking the limit as x->1, since, as you pointed out, the series doesn't strictly converge at x=1), the term in parantheses converges to some number (ln(2) I think), so we still get:

    [tex]2^p \rightarrow 1+Cp[/tex] as [itex]p \rightarrow 0[/itex]

    for some constant C, or in other words:

    [tex]2^{1/n}-1 \rightarrow \frac{C}{n}[/tex] as [itex]n \rightarrow \infty[/itex]

    Since the sum of 1/n diverges, this implies the sum of 2^(1/n)-1 does as well. I think this is all right, but there are some iffy arguments, and the method you posted is probably easier.
     
    Last edited: Feb 2, 2007
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