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Math Help - limit and sequence

  1. #1
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    limit and sequence

    The limit, as n approaches infinity, of (1 + (1/n))^n = e (by e, I mean 2.71828). Use this to find the limit of the following sequence:

    sn = ((n+2)/(n+1))^(n+3)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bearej50 View Post
    The limit, as n approaches infinity, of (1 + (1/n))^n = e (by e, I mean 2.71828). Use this to find the limit of the following sequence:

    sn = ((n+2)/(n+1))^(n+3)
    \begin{aligned}\lim \left(\frac{n+2}{n+1}\right)^{n+3}&= \lim\left(\frac{n+1+1}{n+1}\right)^{n+3}\\ &= \lim\left(1+\frac{1}{n+1}\right)^{n+3}\\&=\lim\lef  t(1+\frac{1}{n+1}\right)^{\left(n+1\right)\left(\f  rac{n+3}{n+1}\right)}\\ &= e^{\lim\frac{n+3}{n+1}}\\&=\boxed{e}\end{aligned}
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bearej50 View Post
    The limit, as n approaches infinity, of (1 + (1/n))^n = e (by e, I mean 2.71828). Use this to find the limit of the following sequence:

    sn = ((n+2)/(n+1))^(n+3)
    note that what you have is s_n = \left( 1 + \frac 1{n + 1} \right)^{(n + 1) + 2} = \left( 1 + \frac 1{n + 1} \right)^2 \cdot \left( 1 + \frac 1{{\color{red}n + 1}} \right)^{\color{red}n + 1}

    now, do you see how to use what you were given?
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