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Comparison test for series.

  1. Apr 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Use the comparison test to show that the series converges, and find the value it converge to by using partial fractions.

    ∑ n=1 -> ∞: [itex]\frac{2}{n^2 + 5n + 6}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    The series can be written as 2 * ∑ n=1 -> ∞: [itex]\frac{1}{n^2 + 5n + 6}[/itex]

    Since 5n + 6 is neglectible:

    2 * ∑ n=1 -> ∞: [itex]\frac{1}{n^2}[/itex]. This series will converge.

    Therefore we can guess that 2* ∑ n=1 -> ∞: [itex]\frac{1}{n^2 + 5n + 6}[/itex] will converge.

    Now I have to find a larger series which also converges.

    [itex]\frac{1}{n^2 + 5n + 6}[/itex] < [itex]\frac{1}{n^2 + 5n}[/itex]

    Now I'm supposed to rewrite [itex]\frac{1}{n^2 + 5n}[/itex] as something + something. This is where I'm stuck.

    Are anyone familiar with this method, and can point me into the right direction?

    Any feedback will as always be appreciated.

    Thanks.
     
  2. jcsd
  3. Apr 28, 2014 #2

    Zondrina

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    Homework Helper

    You did your comparison test properly. ##\frac{1}{n^2}## converges so you know your original series will converge.

    To use partial fractions, write: ##n^2 + 5n + 6 = (n+2)(n+3)##.
     
  4. Apr 28, 2014 #3
    Alright, thank you.

    So, I've rewritten the series as:

    [itex]\frac{1}{n+2}[/itex] - [itex]\frac{1}{n+3}[/itex].

    Adding the series then gives [itex]\frac{1}{3}[/itex] - [itex]\frac{1}{n+3}[/itex]

    Since the limit of [itex]\frac{1}{3}[/itex] - [itex]\frac{1}{n+3}[/itex] = [itex]\frac{1}{3}[/itex]

    The sum which the series converge to is [itex]\frac{1}{3}[/itex].

    It looks correct according to my notes at least.

    Thanks so much for helping me out.
     
  5. Apr 28, 2014 #4

    Zondrina

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    Homework Helper

    No not quite. We have:

    ##n^2 + 5n + 6 = (n+2)(n+3)##.

    So that:

    ##\frac{2}{n^2 + 5n + 6} = \frac{A}{n+2} + \frac{B}{n+3}##

    Where the relationship:

    ##2 = A(n+3) + B(n+2)## must hold.

    If ##n = -3##, then ##B = -2##. If ##n = -2##, then ##A = 2##.

    So really you have ##\frac{2}{n+2} - \frac{2}{n+3}##

    You were missing a factor of ##2##.
     
  6. Apr 28, 2014 #5
    Indeed, my bad. I didn't consider the fact that I had taken it outside of the actual series, and I forgot to multiply it back in to get my final answer. I should be 2 * 1/3 = 2/3.

    Now, it should be good :)
     
  7. Apr 28, 2014 #6

    Mark44

    Staff: Mentor

    Just a little bit more LaTeX will make the summation look a lot nicer.
    Code (Text):
    $$\sum_{n = 1}^\infty \frac{2}{n^2 + 5n + 6}$$
    The above renders as
    $$\sum_{n = 1}^\infty \frac{2}{n^2 + 5n + 6}$$
    It's more convincing to compare directly with ##\sum \frac{1}{n^2}##, a convergent p-series.
    ##\sum \frac{1}{n^2}## is a convergent series, so ##2\sum \frac{1}{n^2}## is, as well.
    It's not hard to show that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##.
     
  8. Apr 28, 2014 #7
    Thanks for the latex code.

    Under normal circumstances, in Calculus 2 (not the actual course, but a equivalent course), would it be needed to actually show that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##? It like 2 < 4. I can't show it or prove it.

    or do you mean that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}## is adequate?

    [itex]\frac{1}{n^2}[/itex] is convergent, 2 * [itex]\frac{1}{n^2}[/itex] is convergent, therefore ##\frac 1 {n^2 + 5n + 6}## is convergent because ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##.
     
  9. Apr 28, 2014 #8

    Zondrina

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    Homework Helper

    It is adequate to write ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##. Although it is easy to show this relationship holds.

    For ##n ≥ 1## it is true that ##n^2 + 5n + 6 > n^2##. Reciprocating the inequality, we obtain ##\frac{1}{n^2 + 5n + 6} < \frac{1}{n^2}##.
     
  10. Apr 28, 2014 #9

    Mark44

    Staff: Mentor

    Whether it's adequate to just write ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}## probably depends on your instructor, and Zondrina's reply shows that it's easy to demonstrate this inequality, working with the reciprocals.

    If your instructor is confident in your ability to justify the inequality, he/she might not require that the work be shown. However, it doesn't hurt to get some practice doing these quick proofs, especially if the problem happened to be slightly more difficult, like this one:
    $$\sum_{n = 1}^\infty \frac 1 {n^2 - n - 6}$$

    The series to compare to is again ##\sum \frac 1 {n^2}##, but this time it's a little harder to show that ##frac{1}{n^2 - n - 6} < \frac 1 {n^2}##.
     
    Last edited by a moderator: May 2, 2014
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