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Covergence/Divergence of an Infinite Series

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Does the infinite series [tex]\sum_{n=1}^{\infty}(\frac{1}{n^(1 + (\frac{1}{n}))})[/tex] converge?


    2. Relevant equations
    Power series [tex]\sum_{n=1}^{\infty}\frac{1}{n^p}[/tex]


    3. The attempt at a solution
    I used the fact that for a power series, if p>1 the series will converge.
    Since [tex]{n^(1+\frac{1}{n})}[/tex] will always be greater than n, the series will converge.

    Is this correct? If not, can someone lead me in the right direction please?
     
  2. jcsd
  3. Mar 18, 2010 #2

    gabbagabbahey

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    What does [tex]\sum_{n=1}^{\infty}\frac{1}{n+1}[/tex] become when you make the substitution [itex]m=n+1[/itex] (and hence [itex]n=m-1[/itex])?
     
  4. Mar 18, 2010 #3
    Sorry if the text isn't clear - the denominator is n^(1 + 1/n).
     
  5. Mar 18, 2010 #4

    gabbagabbahey

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    In that case, your argument is invalid since the exponent depends on the summation index (It's true that [itex]1+\frac{1}{n}\geq 1[/itex], but it isn't a constant value like the [itex]p[/itex] in your power series convergence test).


    You will have to find another convergence test to use instead.
     
  6. Mar 18, 2010 #5
    Sorry but do you have any in mind? The only ones I know are the comparison, ratio, and integral tests, and the function makes these tests impractical (at least I think so).
     
  7. Mar 18, 2010 #6

    gabbagabbahey

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    I think the comparison test is your best bet here.
     
  8. Mar 18, 2010 #7
    I thought about that, as in the power series. Since the exponent on n is between 1 and 2, the series sits right in between a diverging (exponent 1) and a converging (exponent 2) series. Wouldn't that be inconclusive?
     
  9. Mar 18, 2010 #8
    Sorry for the double post, but if I use the limit comparison test, letting the series {a_n} be the series of the question and {b_n} be 1/n, then the limit as n approaches infinity of a_n/b_n would = 1, but since the series {b_n} does not converge, then {a_n} does not converge as a series either. Is that it?
     
  10. Mar 18, 2010 #9

    Gib Z

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    If the ratio exists and is non zero, then a_n and b_n converge or diverge together. You limit is 1 (non zero) so your deduction is correct.
     
  11. Mar 18, 2010 #10
    Thank you guys so much!
     
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