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Thread: [SOLVED] Find the interval of convergence of the series

  1. #1
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    [SOLVED] Find the interval of convergence of the series

    55. $\displaystyle \sum_{n=1}^{\infty} \frac{n^n(x+2)^n}{3^nn!} $

    My Solution

    $\displaystyle \lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} | = \frac{(n+1)^{n+1}(x+2)^{n+1}}{3^{n+1}(n+1)!} * \frac{3^nn!}{n^n(x+2)^n} $

    $\displaystyle = \frac{(n+1)(n+1)^n(x+2)}{3(n+1)n^n} \Longrightarrow \frac{|x+2|}{3} < 1 \Longrightarrow -5 < x < 1 $

    However, the solution from the book goes like this:

    Book Solution

    $\displaystyle \lim_{n\rightarrow\infty} \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}(x+2)^{n+1}} * \frac{n^n(x+2)^n}{3^nn!} $

    $\displaystyle \lim_{n\rightarrow\infty} \frac{3n^n}{(n+1)^n(x+2)} = \frac{3}{x+2} $

    $\displaystyle -\frac{3}{e} -2 < x < \frac{3}{e} -2 $ or

    $\displaystyle -3.104 < x < -0.896 $


    What am I doing wrong? I copied the solution number/letter by number/letter.

    Does the same thing for #56:

    56. $\displaystyle \sum_{n=1}^{\infty} \frac{n!x^n}{5^nn^n} $

    My Solution

    $\displaystyle \lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} | = \frac{(n+1)!x^{n+1}}{(n+1)^{n+1}5^{n+1}} * \frac{5^nn^n}{n!x^n} $

    $\displaystyle \lim_{n\rightarrow\infty} \frac{|x|}{5} < 1 \Longrightarrow -5 < x < 5 $

    Book Solution

    $\displaystyle \lim_{n\rightarrow\infty} \frac{(n+1)^{n+1}5^{n+1}}{(n+1)!x^{n+1}} * \frac{n!x^n}{5^nn^n} $

    $\displaystyle \lim_{n\rightarrow\infty} \frac{(n+1)^n5}{n^nx} = \frac{5}{x} $

    $\displaystyle -5e < x < 5e $ or

    $\displaystyle = -13.591 < x < 13.591 $

    any help is appreciated.
    Last edited by lilaziz1; Apr 7th 2010 at 01:18 PM.
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  2. #2
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    Quote Originally Posted by lilaziz1 View Post
    55. $\displaystyle \sum_{n=1}^{\infty} \frac{n^n(x+2)^n}{3^nn!} $

    My Solution

    $\displaystyle \lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} | = \frac{(n+1)^{n+1}(x+2)^{n+1}}{3^{n+1}(n+1)!} * \frac{3!n!}{n^n(x+2)^n} $

    $\displaystyle = \frac{(n+1)(n+1)^n(x+2)}{3(n+1)n^n} \Longrightarrow \frac{|x+2|}{3} < 1 \Longrightarrow -5 < x < 1 $

    However, the solution from the book goes like this:

    Book Solution

    $\displaystyle \lim_{n\rightarrow\infty} \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}(x+2)^{n+1}} * \frac{n^n(x+2)^n}{3^nn!} $

    $\displaystyle \lim_{n\rightarrow\infty} \frac{3n^n}{(n+1)^n(x+2)} = \frac{3}{x+2} $

    $\displaystyle -\frac{3}{e} -2 < x < \frac{3}{e} -2 $ or

    $\displaystyle -3.104 < x < -0.896 $


    What am I doing wrong? I copied the solution number/letter by number/letter.
    You used $\displaystyle \lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} |$ but it it's supposed to be $\displaystyle \lim_{n\rightarrow\infty} |\frac{a_n}{a_{n+1}} |$, like what the book did... does that help?
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  3. #3
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    Please notice, that
    $\displaystyle \lim\frac{{(n+1)}^n}{n^n}$ = $\displaystyle \lim{(1+\frac{1}{n})}^n$ = e
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  4. #4
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    but isn't the Ratio test $\displaystyle \frac{|a_{n+1}|}{|a_n|} $
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  5. #5
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    :O! i just re-did the problem. thanks a ton zzzoak!
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  6. #6
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    $\displaystyle
    \lim\frac{(n+1)^n}{n^n}\frac{(x+2)}{3} \Longrightarrow \frac{e|x+2|}{3} < 1
    $
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  7. #7
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    Quote Originally Posted by zzzoak View Post
    $\displaystyle
    \lim\frac{(n+1)^n}{n^n}\frac{(x+2)}{3} \Longrightarrow \frac{e|x+2|}{3} < 1
    $
    Yea. Didn't use to pay attention to the n's much til now. thanks once again.
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  8. #8
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    Quote Originally Posted by lilaziz1 View Post
    but isn't the Ratio test $\displaystyle \frac{|a_{n+1}|}{|a_n|} $
    Hmm, you are right, I misread this Wikipedia article and saw that the book started with $\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ and made a wrong conclusion. Sorry about that.
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