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Thread: Convergent series by Ratio Test

  1. #1
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    Convergent series by Ratio Test

    I have to show the following series is either convergent or divergent.

    $\displaystyle \sum_0^\infty \frac{(n^2 +1)^2.2^{2n-1}}{n!}$

    I know this series is convergent by the ratio test, but I am having trouble showing this proof.

    So far I have:

    $\displaystyle \rho = lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = lim_{n\rightarrow\infty} \frac{((n+1)^2 +1)^2.2^{2(n+1)-1}}{(n+1)!} . \frac{n!}{(n^2 +1)^2.2^{2n-1}}$

    I simplified to
    $\displaystyle \rho = lim_{n\rightarrow\infty} \frac{((n+1)^2 +1)^2}{(n^2+1)^2}.\frac{2^{2n+1}}{2^{2n-1}}.\frac{n!}{(n+1)n!}}$
    $\displaystyle = lim_{n\rightarrow\infty} \frac{(n^2 +2n+2)^2}{(n^2+1)^2}.2^2.\frac{1}{(n+1)}$
    $\displaystyle = lim_{n\rightarrow\infty} \frac{4.(n^2 +2n+2)^2}{(n^2+1)^2(n+1)}$

    This is where I am stuck.

    Could someone help me check if this is right so far and help me with what comes next?
    Last edited by tammyl; May 21st 2013 at 03:22 AM.
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    Re: Convergent series by Ratio Test

    It looks ok to me so far. It might be a good idea to expand all your brackets and then divide top and bottom by the highest power of n...
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    Re: Convergent series by Ratio Test

    Thanks for the response.

    So, expanding the brackets:
    $\displaystyle \rho=lim_{n\rightarrow\infty} \frac{4n^4 + 16n^3+32n^2+32n+16}{n^5+n^4+2n^3+2n^3+n+1}$

    dividing through by $\displaystyle n^5$ gives:

    $\displaystyle \rho=\frac{0}{1}=0<1$

    therefore series is convergent.

    Is this an appropriate way to complete the problem?
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    Re: Convergent series by Ratio Test

    You need to show more working out, but yes, your logic is correct
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    Re: Convergent series by Ratio Test

    It converges ..
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    Re: Convergent series by Ratio Test

    Quote Originally Posted by tammyl View Post
    $\displaystyle = lim_{n\rightarrow\infty} \frac{4.(n^2 +2n+2)^2}{(n^2+1)^2(n+1)}$
    If you were to multiply the terms in numerator out you would see that the numerator has highest power of $\displaystyle n^4$ and the denominator has highest power $\displaystyle n^5$
    That means that the denominator will increase faster than the numerator and the limit is 0.
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