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Convergent/Divergent, find sum

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series is convergent or divergent. If it is convergent, find its sum.
    [tex]\sum\limits_{n=1}^{\infty} (\frac{1}{e^n}+\frac{1}{n(n+1)})[/tex]


    2. Relevant equations



    3. The attempt at a solution
    So I found it's convergent:
    [tex]\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{(n^{2}+n)})=\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})[/tex]
    [tex]\lim_{n \to \infty}((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})=0+0+0[/tex]
    ∴ the sum is convergent.

    Now how would I find the sum of the series?
     
  2. jcsd
  3. Mar 25, 2013 #2

    Dick

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    Having a limit of 0 doesn't show it's convergent. (1/e)^n is geometric. The other two terms from a telescoping series. Find the sums separately.
     
  4. Mar 25, 2013 #3
    My professor said that a limit as n approaches infinity that equals zero means the series is convergent.. And I see what you're saying about the sum. Now I'm just confused on the convergent part.
    Edit:
    For the (1/e)^n I guess you could also say that since |1/e|<1, the series is convergent.
     
  5. Mar 25, 2013 #4

    Dick

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    Right for the (1/e)^n. And I hope that's not really what your professor said. If the limit is NOT zero, then it's NOT convergent. Approaching zero doesn't mean it converges. 1/n approaches zero but the series 1/n is not convergent.
     
  6. Mar 25, 2013 #5
    Oh I see the problem here. 1/n is a harmonic series and divergent. Now how would I go about proving it is convergent?
     
  7. Mar 25, 2013 #6
    Also I just checked me notes it says the following:
    If [itex]\sum_{n=1}^{\infty} a_{n}[/itex] is convergent, then [itex]\lim_{n \to \infty}a_{n}=0[/itex] and the opposite if an does not equal 0.

    I think there is a difference in what I said and what my notes say right?
     
  8. Mar 25, 2013 #7

    Dick

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    You know that the (1/e)^n part is convergent and you know it's sum. Now do the 1/n-1/(n+1) part. Like I said, it telescopes. If you know those two parts converge then the sum converges.
     
  9. Mar 25, 2013 #8

    Dick

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    Yes, what you quoted doesn't say that if the limit is 0 then the sum converges.
     
  10. Mar 25, 2013 #9
    Ok then I did it correctly.
    Thank you, again.
     
  11. Mar 25, 2013 #10
    So you cannot say the opposite, if [itex]\lim_{n \to \infty} a_{n}=0[/itex] then [itex]\sum_{n=1}^{\infty}a_{n}[/itex] converges.?
     
  12. Mar 25, 2013 #11

    Dick

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    Doesn't the example of ##a_n=1/n## show that you can't?
     
  13. Mar 25, 2013 #12
    I mean are there any other examples of this, other than the harmonic series?
     
  14. Mar 25, 2013 #13

    Dick

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    Many. 1/n^(1/2) doesn't converge. 1/log(n) doesn't converge. log(n)/n doesn't converge. But the limits are all zero.
     
  15. Mar 25, 2013 #14
    Alright, thanks a lot.
     
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