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Math Help - series convergency

  1. #1
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    series convergency

    Hello, I am looking at the following series:

    sum, n=1 to infinity
    {n^3} / e^n

    using the ratio test, we get to:
    {(n+1)^3} / en^3
    however I cannot understand the final step which goes to 1/e which less than 1, so proves convergency.

    Can someone please show me how it goes from :
    {(n+1)^3} / en^3
    to
    1/e

    Thanks kindly.
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  2. #2
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    Re: series convergency

    Quote Originally Posted by fran1942 View Post
    sum, n=1 to infinity
    {n^3} / e^n
    \lim_{n\to\infty}\left|\frac{(n+1)^3/e^{n+1}}{n^3/e^n}\right|

    =\lim_{n\to\infty}\frac{(n+1)^3e^n}{n^3e^{n+1}}

    =\frac1e\lim_{n\to\infty}\frac{(n+1)^3}{n^3}

    =\frac1e\lim_{n\to\infty}\left(\frac{n+1}n\right)^  3

    =\frac1e\lim_{n\to\infty}\left(1+\frac1n\right)^3

    =\frac1e < 1
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