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Find convergence of series

  1. Apr 29, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    [tex]\sum_{n=0}^{\infty}\frac{1}{n^2+3n+2}[/tex]

    The attempt at a solution
    I'm wondering if there is only one way of solving this?

    Here is what i've done: First, converting into partial fractions. Is there a way to do it without converting to partial fractions?
    [tex]\sum_{n=0}^{\infty}\frac{1}{n^2+3n+2}=\sum_{n=0}^{\infty}\frac{-1}{n+2}+\sum_{n=0}^{\infty}\frac{1}{n+1}[/tex]
    [tex](u1)_n=\frac{-1}{n+2}[/tex]Then, for large n,[tex](v1)_n=\frac{-1}{n}
    \\(u2)_n=\frac{1}{n+1}[/tex]Then, for large n,[tex](v2)_n=\frac{1}{n}[/tex]
    Initially, i decided to use the comparison test, but it failed (somewhat). Maybe i did it wrong?

    Considering [itex]\sum_{n=0}^{ \infty}\frac{-1}{n+2}[/itex]: if i ignore the negative sign, then [itex]0<(u1)_n\leq (v1)_n[/itex]. But [itex](v1)_n[/itex] diverges, so the test does not apply in this case.
    Now, if i were to consider the negative sign and put [itex](u1)_n[/itex] and [itex](v1)_n[/itex] on a number line, then i can see that [itex]0<(v1)_n\leq (u1)_n[/itex], in which case, the comparison test works and the series diverges. But this is an awkward conclusion in my opinion. So, which of these methods is the correct one?

    Considering [itex]\sum_{n=0}^{\infty}\frac{1}{n+1}[/itex]: [itex]0<(u2)_n\leq (v2)_n[/itex], so the comparison test fails since [itex](v2)_n[/itex] diverges.

    So, i opted for the limit comparison test and both of the sum series diverge, therefore the original series also diverges. It's unusual, as the comparison test failed, yet the limit comparison test agrees?

    If what i've done above is correct, then i don't understand why this question is set before the comparison test is even introduced in my book. In fact, this problem is right after the geometric series, so i should probably use the latter to solve for this problem, but i can't figure out how to do it.
     
    Last edited: Apr 29, 2012
  2. jcsd
  3. Apr 29, 2012 #2

    tiny-tim

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    hi sharks! :smile:
    hint: what is [itex]\sum_{n=0}^{N}\frac{-1}{n+2}+\sum_{n=0}^{N}\frac{1}{n+1}[/itex] ? :wink:
     
  4. Apr 29, 2012 #3

    sharks

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    Using the Integral test? But that method has been introduced much later in the book, after this problem. So, i think i should solve it using geometric series or just apply the theorems that i use for sequences (although that would be incorrect, i think):
    [tex]\sum_{n=0}^{ \infty}\frac{-1}{n+2}+\sum_{n=0}^{\infty}\frac{1}{n+1}=0[/tex]
    As n approaches infinity, the values of each of these sums become 0. But i'm not sure about this, since i'm using a sequence theorem to solve series.
     
    Last edited: Apr 29, 2012
  5. Apr 29, 2012 #4
    If we say [tex] n^2 < n^2+3n+2 [/tex]

    So

    [tex] \frac{1}{n^2} > \frac{1}{n^2+3n+2} [/tex] and

    [tex] \sum \frac{1}{n^2} [/tex] converges so by the comparison test it converges ?

    Would that be wrong ?


    PS: NOT GIVING A SOLUTION HE SAID HE TRIED THIS SO I JUST WANT TO SEE IF MY WORKING IS OKAY
     
  6. Apr 29, 2012 #5

    sharks

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    Well, your reasoning would work; according to the comparison test, it would converge, but then you haven't expressed the expression in terms of partial fractions. I think for this type of problem, you must break it down into partial fractions before trying any test.
     
  7. Apr 29, 2012 #6
    Is there a particular reason why you want to use partial fractions ? Seems like they would complicate things unnecessarily ?
     
  8. Apr 29, 2012 #7

    sharks

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    That's what i would do if i were to integrate that expression, and since i should be able to use the Integral test, i think that's the way to go. However, i'm not even supposed to be using the Comparison or Integral Test, as it's not introduced in the book yet, since this problem is at the beginning of the chapter.
     
  9. Apr 29, 2012 #8

    sharks

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    If it was a geometric series, then i could have found the limit.

    OK, so if i factorize n from the denominator and numerator and then take the limit:
    [tex]\sum_{n=0}^{N}\frac{-1}{n+2}+\sum_{n=0}^{N}\frac{1}{n+1}=\lim_{n \to \infty}\frac{-1/n}{1+2/n}+\lim_{n \to \infty}\frac{1/n}{1+1/n}=0+0=0[/tex] Is that correct?

    Was it necessary to find the partial fractions? And what if i had a quadratic in the denominator, that was not factorisable?
     
    Last edited: Apr 29, 2012
  10. Apr 29, 2012 #9

    tiny-tim

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    what is -(1/3 + 1/4 + … 1/11 + 1/12) + (1/2 + 1/3 + … + 1/10 + 1/11) ? :wink:
     
  11. Apr 29, 2012 #10

    sharks

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    [tex]\frac{1}{2}-\frac{1}{12}[/tex]
    Ah yes, i see your point...
    [tex]\sum_{n=0}^{\infty}\frac{1}{n^2+3n+2}=\sum_{n=0}^{ \infty}\frac{-1}{n+2}+\sum_{n=0}^{\infty}\frac{1}{n+1}=\sum^{ \infty }_{n=0} \left(1-\frac{1}{n+2}\right)[/tex]
    [tex]\sum^{ \infty }_{n=0} \left(1-\frac{1}{n+2}\right)=\sum^{ \infty }_{n=0} 1-\sum^{ \infty }_{n=0} \left(\frac{1}{n+2}\right)=1-\sum^{ \infty }_{n=0} \left(\frac{1}{n+2}\right)[/tex] Then, i don't know. I'm back to the fraction form.
     
    Last edited: Apr 29, 2012
  12. Apr 29, 2012 #11

    vela

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    That argument is fine to prove the series converges, but by using partial fraction, you can also figure out what the series converges to.
     
  13. Apr 29, 2012 #12

    tiny-tim

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    nooo :wink:

    leave out a ∑ :smile:
     
  14. Apr 29, 2012 #13

    sharks

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    Oh! You are correct.
    [tex]\sum_{n=0}^{\infty}\frac{1}{n^2+3n+2}=\sum_{n=0}^{ \infty}\frac{-1}{n+2}+\sum_{n=0}^{\infty}\frac{1}{n+1}=\left(1-\frac{1}{n+2}\right)[/tex]
    So, i guess that's the finally answer? Or do i need to replace n with ∞, thereby getting the final answer=1?
     
  15. Apr 29, 2012 #14

    vela

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    That's not correct. You can't actually break it into two series like that. What you can say is that
    $$\frac{1}{n^2+3n+2} = \frac{1}{n+1}-\frac{1}{n+2}$$ (note there are no summations in what I just wrote) so that
    $$\sum_{n=0}^\infty \frac{1}{n^2+3n+2} = \sum_{n=0}^\infty \left(\frac{1}{n+1}-\frac{1}{n+2}\right)$$ You can't break the righthand side into the difference of two summations because neither ##\displaystyle \sum_{n=0}^\infty \frac{1}{n+1}## nor ##\displaystyle\sum_{n=0}^\infty \frac{1}{n+2}## converges.

    Take a look back to post #2 and note that Tiny Tim used N for the upper limit, not ∞. In that case, you have sums of finite number of terms, so you can break up the original sum into two separate sums. Specifically, that's the partial sum sN, whose limit as N goes to infinity is what the series, by definition, converges to. What is the partial sum sN equal to?
     
  16. Apr 29, 2012 #15

    tiny-tim

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    yup! :biggrin:

    (btw, this is quite a common technique in exam problems …

    you need to be familiar with it, and to look out for it! :wink:)​
     
  17. Apr 29, 2012 #16

    sharks

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    [tex]S_N=\left(1-\frac{1}{n+2}\right)?[/tex]
    Honestly, i think i'm in big trouble with differentiating between series and sequences, and how to figure them out without incorrectly mixing different techniques and theorems. It's not so obvious to me. Like mixing two types of noodles. I'll need to go over the Infinite Series chapter again.
     
  18. Apr 29, 2012 #17

    vela

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    The little n should be a capital N on the righthand side, but otherwise it's correct. You can see the limit of the sequence sN is 1, so the series converges to 1.

    It's a bit confusing because there are two sequences associated with a series. First, there's the sequence an that make up the terms of the series. This sequence has to converge to 0 for the series to converge. Second, there's the sequence sn of partial sums. When you say an infinite series converges, it means the sequence of partial sum converges.
     
  19. Apr 29, 2012 #18

    sharks

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    Thank you, vela. That helped a lot. I wonder why my books don't summarize it that way at the end of the chapter.:rolleyes:

    But, what if an does not converge to 0? Suppose an converges to 1. Then, what conclusions can be drawn?

    Edit: OK, i think i finally understand. If an does not converge to 0, then according to nth term test for divergence, the series is guaranteed to diverge.
     
    Last edited: Apr 29, 2012
  20. Apr 29, 2012 #19

    LCKurtz

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    See if your text doesn't have a section on telescoping series.
    Probably the first theorem on infinite series says if ##\sum_{n=1}^\infty a_n## converges, then ##a_n\rightarrow 0##. The contrapositive of this statement is that if ##a_n## does not go to zero, the series diverges. The nth term of a convergent series must go to zero. Otherwise the series diverges. But note that the nth term going to zero is only a necessary condition for convergence of the series, it isn't sufficient. If it were sufficient you wouldn't need all those convergence tests. All you would have to check is whether the nth term goes to zero. Too bad life isn't that simple.
     
  21. Apr 29, 2012 #20

    sharks

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    In my book, the only example on telescoping series deals with partial sums. I didn't understand any of that before, as i just crammed over it. But now i appreciate that there are 2 different kinds of sequences associated with series. That little piece of crucial information is unfortunately not mentioned anywhere as i've reviewed that chapter several times.

    But i have a doubt: for the series to converge, the sequence an must converge to 0, but does that mean that the series converges to 0?
     
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