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Alternating serries

  1. Aug 23, 2009 #1
    [tex]\sum_{n=1}^{\infty}\frac{(-10)^n}{n!}[/tex]
    im seeing the terms increasing here DIVERGENT
    [tex]\sum_{n=1}^{\infty}\frac{10^n}{n!}[/tex]
    [tex]\lim_{n->\infty}\sqrt[n]{\frac{10^n}{n!}}>1 [/tex]
    text is showing absolute convergence
    [tex]\sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2} [/tex]
    [tex]\lim_{n->\infty}\frac{n^2+3}{(2n-5)^2}\neq0[/tex]
    divergent
    [tex]\frac{n^2+3}{(2n-5)^2}\leq1[/tex]
    [tex]\sum_{n=1}^{\infty}1 -> \infty[/tex]
    [tex]\lim_{n->\infty}{\frac{n^2+3}{(2n-5)^2}>0[/tex]
    divergent by limit comparison test

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  2. jcsd
  3. Aug 23, 2009 #2

    zcd

    User Avatar

    You might want to try ratio test (for absolute convergence) in the case of factorials.
    [tex]\lim_{n\to\infty}\left|\frac{10^{n+1}}{(n+1)n!}\cdot\frac{n!}{10^{n}}\right|=\lim_{n\to\infty}\left|\frac{10}{n+1}\right|=0[/tex]
     
  4. Aug 23, 2009 #3
    can someone please confirm the method used to solve the second
     
  5. Aug 23, 2009 #4
    If the second one is:

    [tex]
    \sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2}
    [/tex]

    Then you should be fine with the nth term test:

     
    Last edited by a moderator: May 4, 2017
  6. Aug 23, 2009 #5

    zcd

    User Avatar

    If you're testing for conditional convergence, then the conditions are [tex]a_{n}[/tex] must be constantly decreasing (monotonic decreasing) and [tex]\lim_{n\to\infty}a_{n}=0[/tex]. In the case of the second problem, the test you took would apply for both conditional and absolute convergence, so it's right.
     
  7. Aug 24, 2009 #6
    For this one: [tex]\sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2}[/tex], you might want to be a little bit more careful about your argument; you've written a couple things down, and some are not entirely correct (but could be just the awkward formatting that's making things hard to read).

    It seems to me you're confusing a few concepts here. You wrote [tex]\frac{n^2+3}{(2n-5)^2} \leq 1[/tex] and then wrote that [tex]\sum_{n=1}^{\infty}1 \to \infty[/tex] --- are you attempting to use the Comparison Test here? Because knowing that [tex]\frac{n^2+3}{(2n-5)^2} \leq 1[/tex] holds and even knowing that [tex]\sum_{n=1}^{\infty}1 \to \infty[/tex] says nothing about the convergence of [tex]\sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2}[/tex]. Besides, the inequality you wrote down is wrong. Set [tex]n = 2[/tex] and you immediately have [tex]\frac{2^2 + 3}{(2\cdot2 - 5)^2} = \frac{4 + 3}{(-1)^2} = 7 \not\leq 1[/tex].

    A good approach here would be to use the Alternating Series Test.
     
  8. Aug 24, 2009 #7
    Incorrect. Just for fun, can you do this limit correctly?
     
  9. Aug 24, 2009 #8
    [tex]\lim_{n->\infty}\frac{10}{(n!)^(1/n)} = 10 [/tex]
    I assumed
     
  10. Aug 24, 2009 #9
    [itex]n![/itex] is much bigger than [itex]10^n[/itex] when [itex]n[/itex] is large.

    For example:

    [tex]10000000000000000000000000000000000000000 = 10^{40}[/tex]

    [tex]815915283247897734345611269596115894272000000000 = 40! = 8.159 \times 10^{47}[/tex]
     
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