1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinite Series Test (Ratio Test get 1)

  1. Sep 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Test if the infinite series converge or diverge.

    2. Relevant equations
    [tex]
    \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}
    [/tex]


    3. The attempt at a solution
    I tried Ratio test:
    [tex]a_{n+1} = \frac{4n+7}{(n+1)(n+2)(n+3)}[/tex]
    [tex]a_{n} = \frac{4n+3}{n(n+1)(n+2)}[/tex]
    [tex]\left|\frac{a_{n+1}}{a_{n}}\right| = \frac{4n+7}{(n+1)(n+2)(n+3)} \times \frac{n(n+1)(n+2)}{4n+3}
    = \frac{n(4+7n)}{(n+3)(4n+3)}
    = \frac{4n^{2}+7n}{4n^{2}+15n+9}[/tex]
    [tex]lim_{n\rightarrow\infty} \left|\frac{a_{n+1}}{a_{n}}\right| = lim_{n\rightarrow\infty} \frac{4+\frac{7}{n}}{4+\frac{15}{n}+\frac{9}{n^{2}}} = \frac{4}{4} = 1[/tex]
    The answer is inconclusive, and I can't seem to think of any other test yet.
    Anyone can help me with this?
    I will much appreciate it. Thanks!
     
    Last edited: Sep 2, 2010
  2. jcsd
  3. Sep 2, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Think about what the terms look like for large n. Try a comparison test.
     
  4. Sep 2, 2010 #3
    Thanks for your help.
    This is what I done:
    [tex]
    n^{3} > n(n+1)(n+1)[/tex]
    [tex]
    \frac{4n+3}{n^{3}} < \frac{4n+2}{n(n+1)(n+2)}[/tex]
    [tex]
    \frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}[/tex]
    Both converge

    Therefore, [tex]\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}[/tex] converges.


    Is that correct?
     
    Last edited: Sep 2, 2010
  5. Sep 2, 2010 #4
    How is n^3 > n^3 + 2n^2 + n?
     
  6. Sep 2, 2010 #5
    Isn't this the same with the [tex]

    n^{3} > n(n+1)(n+1)
    [/tex] above?
     
  7. Sep 2, 2010 #6

    Mark44

    Staff: Mentor

    Yes, so why do you think that n3 > n(n + 1)(n + 2)?
     
  8. Sep 2, 2010 #7
    Opps, sorry,
    found the careless mistake.

    should be
    n(n+1)(n+2) > n3

    so
    [tex]\frac{4n+3}{n(n+1)(n+2)} < \frac{4n+3}{n^{3}}[/tex]
    Thanks for pointing me out

    [tex]\frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}[/tex]
    Both converge

    Therefore, [tex]\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}[/tex] converges.
     
  9. Sep 2, 2010 #8
    Go further with your inequalities:
    (use the term by term comparison test)

    [tex]
    \frac{4n+3}{n(n+1)(n+2)}\leq \frac{7n}{n^{3}}= 7\frac{1}{n^{2}}
    [/tex]
     
  10. Sep 2, 2010 #9
    How do you get the 7n?
     
  11. Sep 2, 2010 #10

    Mark44

    Staff: Mentor

    4n + 3 <= 7n for all n >= 1
     
  12. Sep 2, 2010 #11
    Got it!
    Thanks!
     
  13. Sep 2, 2010 #12
    So, the final answer should look like this:
    [tex]
    \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}
    [/tex]

    n(n+1)(n+2) [tex]\geq[/tex] n3
    [tex]\frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}}[/tex]

    4n+3 [tex]\leq[/tex] 7n , for all n [tex]\geq[/tex] 1

    [tex]\frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}}[/tex]

    [tex]\sum_{n=1}^{\infty}\frac{7n}{n^{3}} = \sum_{n=1}^{\infty}\frac{7}{n^{2}}[/tex]
    Converge P-series (p > 1)

    According to Comparison test,
    since [tex]\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} \leq \sum_{n=1}^{\infty}\frac{7}{n^{2}}[/tex]
    [tex]and \sum_{n=1}^{\infty}\frac{7}{n^{2}} converges[/tex],
    [tex]therefore
    \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} converges.
    [/tex]
     
  14. Sep 2, 2010 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Very nice.
     
  15. Sep 2, 2010 #14
    Thanks,
    and thanks for everyone that helps.
    :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Infinite Series Test (Ratio Test get 1)
  1. Ratio Test Of A Series (Replies: 3)

Loading...