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Math Help - Need help with these convergence Tests?

  1. #1
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    Need help with these convergence Tests?

    Do these converge absolutely, conditionally, or diverge?

    sigma from n = 1 to infinity of (n + 1)^2 / n!

    sigma from n = 1 to infinity of (1/ (arctan(n))^(2n)

    explain please!
    Last edited by Playthious; November 28th 2010 at 07:48 PM.
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  2. #2
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    I first break it up. (n+1)^2=n^2+2n+1

    Now you can evaluate three separate Summations.

    I think the 2nd one can be evaluate like a geometric since ArcTan(Infinity)=pi/2
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  3. #3
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    For the first, try the ratio test.

    Evaluate \displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|. If this is \displaystyle <1 the series converges absolutely, if this is \displaystyle >1 the series diverges. If this \displaystyle =1 the test is inconclusive.
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  4. #4
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    so for the first one I applied the ratio test and got
    lim (as n goes to infinity) of (n^2 + 2n + 1) / ( (n+1)(n^2) ) + lim (as n goes to infinity) (2n + 1)/( (n+1)(n^2) + lim (as n goes to infinity) of 1/(n+1)
    and since all of these go to zero then the sum converges absolutely?

    and for the second one you have
    sum (from 1 to infinity) 1 / (pi/2)^2n
    and because pi/2 is greater than 1 it just diverges by geometric series test?
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  5. #5
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    You are mixing up two separate suggestions. I thought you wanted to solve the series but you just want to know if they converge. Just follow Post #3

    \displaystyle a_{n+1}=\frac{(n+2)^2}{(n+1)!}

    \displaystyle \frac{\frac{(n+2)^2}{(n+1)!}}{\frac{(n+1)^2}{n!}}\  rightarrow \frac{(n+2)^2}{(n+1)^3}\rightarrow \frac{n^2}{n^3}\rightarrow\lim_{n\to\infty}\frac{1  }{n}=0
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  6. #6
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    ahh i see
    so lim (as n goes to infinity) (n+2)^2 / ( (n+1)(n+1)2 which is just 0 so it converges abs

    what about the second one?
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  7. #7
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    Try the nth root test
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