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Diverging and converging infinite series

  1. Sep 16, 2010 #1
    I looked through some tutorials to find intervals of divergence and tests for divergence...
    My series:
    [PLAIN]http://img843.imageshack.us/img843/4193/51453212.jpg [Broken]
    a and x are constants...
    I did the ratio test and i get [tex]\rho[/tex]=1, so i tried to apply the limit test to see if an is zero or does not exist... The problem is, i am not sure about my method...
    Do i divide the numerator(1) by n also, if so, the test gives an=0..... But then again i am not sure...
    I learned about these tests from http://www.math.unh.edu/~jjp/radius/radius.html"...

    Could someone please tell me if this sum is convergent or divergent (and the method you used to find this out)

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 16, 2010 #2
  4. Sep 16, 2010 #3


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    Take x outside the summation. The terms in the sum behave like 1/n for large n. This series is well known divergent. The sum behaves like ln(n).
  5. Sep 21, 2010 #4
    That makes sense, thanks.. I have one more question...

    I was fiddling with my calculator and typed "Ans-(Ans-[tex]\sqrt{Ans}[/tex]-2)" and kept clicking =...(The expression is in the same form as xn - f(x)/f'(x) but is just equal to [tex]\sqrt{Ans}[/tex]+2....
    I found that it converges as 4. It seemed logical enough....But then i decided to form the equation that represents f(x)...

    so basically, what i got out of the expression is that y/(dy/dx)=x-[tex]\sqrt{x}[/tex]-2
    Integrating(painfully), ln y=k(x-[tex]\sqrt{x}[/tex]-2)([tex]\sqrt{x}[/tex]-2/[tex]\sqrt{x}[/tex]+2)1/3

    Its root x -2 in the numerator not root 3 -2...

    Where k is a constant... I should have tried with something simpler, but is this the correct way to form an approximation of a function that converges using Newton's method (or am i just being an oaf)
    If it is the former, then how do i find k.
    If it is the latter, then how do i find f(x) for :
    [PLAIN]http://img714.imageshack.us/img714/5351/96407726.png [Broken]

    Last edited by a moderator: May 4, 2017
  6. Sep 24, 2010 #5
    Sorry, it is not summation to infinity, but rather term at infinity for the last expression.
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