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Homework Help: Another sum of series quesiton

  1. Apr 24, 2005 #1
    Consider the series

    [tex]\sum_{n=1}^\infty \frac{n}{\sqrt{5n^2+5}}[/tex]

    Value ______

    [tex]a_1 = .316227766, a_2 = 2/5, a_3 = .4242640687, a_4 = .4338609156[/tex]

    there doesnt seem to be any common ratio, so that means that this isnt a geometric series right?

    well i think i can simplify the equation to:

    [tex]\frac{1}{\sqrt{5}}\sum_{n=1}^\infty \frac{n}{\sqrt{(n^2+1)}}[/tex]

    hmm, that's as far as i got, can someone help me ?
    Last edited: Apr 24, 2005
  2. jcsd
  3. Apr 24, 2005 #2
    Reevaluate [itex]a_2[/itex]
  4. Apr 24, 2005 #3
    i made the changes, but i still dont see a common ratio
  5. Apr 24, 2005 #4
    [tex]\frac{1}{\sqrt{5}}\sum_{n=1}^\infty \frac{n}{\sqrt{(n^2+1)}}[/tex]

    That thing is only going to have a value if it's convergent, correct? I don't really remember much about this stuff.

    If a series converges to a finite value, its sequence a must converge to zero.

    Contrapositively, if the sequence a does not converge to zero, the series does not converge to a finite value.

    Here, the sequence a is
    which converges to unity, not zero, as n approaches infinity.
    This implies that the series in question is divergent.

  6. Apr 24, 2005 #5
    it converges for sure, it was the first question asked.
  7. Apr 24, 2005 #6


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    It does not converge ILoveBaseball; you must have mistyped.

    If [tex]a_{n}=\frac{n}{\sqrt{5n^{2}+5}}[/tex]
    But a necessary requirement for convergence of the series [tex]\sum_{n=1}^{\infty}a_{n}[/tex] is that we have [tex]\lim_{n\to\infty}a_{n}=0[/tex]
  8. Apr 24, 2005 #7
    What reason do you have to think it does converge? Prove it mathematically.
  9. Apr 24, 2005 #8
    sorry, you guys are right. for some reason there was a glitch, even if i selected converge, i got 50% of the problem correct(which means i got the first question right). but if i selected diverge, i got 100% of the problem(two problems = 100%) correct.
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