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Determine if series is con/divergent

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Ʃ[√(n+1)-√(n)]/√(n^2 +n)

    3. The attempt at a solution

    well i split the series up first.

    Ʃ√(n+1)/√(n^2 +n) - Ʃ√(n)/√(n^2 +n)

    next:

    Ʃ[(n+1)/(n^2 +n)]^1/2 - Ʃ[n/(n^2 +n)]^1/2

    at this point im not sure if i can take the limit and divide all the terms by the highest power of n.
     
  2. jcsd
  3. May 12, 2012 #2

    Dick

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    Both of the series you've split it up into diverge. They are like 1/sqrt(n). I'd suggest you multiply numerator and denominator by sqrt(n+1)+sqrt(n) first. It's the quadratic conjugate trick. Then use a comparison test.
     
    Last edited: May 12, 2012
  4. May 13, 2012 #3
    so im basically using the comparison test? im comparing these series to 1/n, which is divergent, and because these series are less than 1/n it also is divergent. also considering that 1 of the now 2 series is diveregent the whole thing is divergent.
     
  5. May 13, 2012 #4

    Dick

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    The two series you've split it up into diverge. The original series does not diverge. Read my post again.
     
    Last edited: May 13, 2012
  6. May 13, 2012 #5

    sharks

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    If you split up the following fraction into its sum;
    [tex]\sum^{\infty}_{n= ?}\frac {\sqrt{n+1}-\sqrt n}{\sqrt {n^2 +n}}=\sum^{\infty}_{n= ?}\frac {\sqrt{n+1}}{\sqrt {n^2 +n}}-\sum^{\infty}_{n= ?}\frac {\sqrt n}{\sqrt {n^2 +n}}[/tex]
    and then take the limits for both, you will get zero in each case, which by the nth-term test for divergence, doesn't say if both series converge or diverge.
    Normally, isn't the multiplication of the conjugate done with the denominator, instead of the numerator as you suggested?
     
    Last edited: May 13, 2012
  7. May 13, 2012 #6

    LCKurtz

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    There isn't any "normally". You can rationalize either a numerator or denominator. What you do depends on your problem and what your goal with the particular problem is. Dick's post is spot on for a good suggestion.
     
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