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Math Help - Infinite Series

  1. #1
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    Infinite Series

    The Problem:
    summation as 'n' goes from 2 to infinity of n^20/2^n .

    The ans says it converges to 1/2 but cant see how it does this since the terms seem to be increasing in magnitude when you plug numbers in.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by heatly View Post
    The Problem:
    summation as 'n' goes from 2 to infinity of n^20/2^n .

    The ans says it converges to 1/2 but cant see how it does this since the terms seem to be increasing in magnitude when you plug numbers in.
     \displaystyle \sum _{n=2} ^{\infty} \frac {n^{20}}{2^n}

    so you have that your a_n is

    \displaystyle  a_n =  \frac {n^{20}}{2^n}

    now apply some of the tests for convergence ...

    you can use Cauchy's test of convergence :

     \displaystyle q= \lim _{n\to \infty}  \sqrt [n] {a_n} = \lim_{n \to \infty }\sqrt [n] { \frac {n^{20}}{2^n}} = \left\{\begin{matrix}<br />
 q<1 & converges  \\ <br />
 q>1 & diverges \\ <br />
 q=1 & ???  <br />
\end{matrix}\right.



    or with D'Alambert's test for convergence :

     \displaystyle q= \lim_{n \to \infty } \frac {a_{n+1}}{a_n} =\lim_{n \to \infty } \frac {\frac {(n+1)^{20}}{2^{n+1}} }{\frac {n^{20}}{2^n}} =  \left\{\begin{matrix}<br />
 q<1 & converges  \\ <br />
 q>1 & diverges \\ <br />
 q=1 & ???  <br />
\end{matrix}\right.


    or with Raabe's test of convergence :
     \displaystyle q= \lim_{n \to \infty } n(\frac {a_{n}}{a_{n+1} }-1 )  =\lim_{n \to \infty } n (  \frac {\frac {n^{20}}{2^n} }{\frac {(n+1)^{20}}{2^{n+1}}}-1 ) =  \left\{\begin{matrix}<br />
 q>1 & converges  \\ <br />
 q<1 & diverges \\ <br />
 q=1 & ???  <br />
\end{matrix}\right.


    or you can show by the definition that sum converges or diverges (with partial sums .... )
    Last edited by yeKciM; September 9th 2010 at 02:56 AM.
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by heatly View Post
    The Problem:
    summation as 'n' goes from 2 to infinity of n^20/2^n .

    The ans says it converges to 1/2 but cant see how it does this since the terms seem to be increasing in magnitude when you plug numbers in.
    The first term of the series is (2^20)/4 = 2^18. So the claim that the series converges to a value of 1/2 is absolute rubbish.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by mr fantastic View Post
    The first term of the series is (2^20)/4 = 2^18. So the claim that the series converges to a value of 1/2 is absolute rubbish.
     \displaystyle \sum _{n=2} ^{\infty} \frac {n^{20}}{2^n}

    so based on D'Alambert test of convergence :

    \displaystyle q= \lim_{n \to \infty } \frac {a_{n+1}}{a_n} =\lim_{n \to \infty } \frac {\frac {(n+1)^{20}}{2^{n+1}} }{\frac {n^{20}}{2^n}} =\lim_{n \to \infty } \frac {\frac {(n+1)^{20}}{2^n\cdot 2}}   {\frac {n^{20}}{2^n}}  = \lim_{n \to \infty } \frac {n^{20 } + ... } { 2n^{20} } = \frac {1}{2}

    now because  q = \frac {1}{2} <1 based on D'Alambert test of convergence sum  \displaystyle \sum _{n=2} ^{\infty} \frac {n^{20}}{2^n} converges

    maybe that's what ans was about
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