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Math Help - Converging sequence

  1. #1
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    Converging sequence

    Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges, find its limit.

    a_n = \left(1 + \frac{1}{n^2}\right)^n


    I was never very good at limits. I know that the limit of \left(1 + \frac{1}{n}\right)^n as n goes to infinity is e, but only because they made us memorize it. I don't understand why that is, or how to use it to solve this problem. About the only technique I know you can use is the squeeze theorem, but I never understood how to come up with the two squeezing functions. Could someone explain how to do this problem?

    Thank you!
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  2. #2
    Super Member girdav's Avatar
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    Re: Converging sequence

    Hint: take logarithm and use for x\geq 0: 0\leq \ln(1+x)\leq x.
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  3. #3
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    Re: Converging sequence

    let y = \lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n

    \ln{y} = \lim_{n \to \infty} \ln \left(1 + \frac{1}{n^2}\right)^n

    \ln{y} = \lim_{n \to \infty} n \cdot \ln \left(1 + \frac{1}{n^2}\right)

    \ln{y} = \lim_{n \to \infty} \frac{\ln \left(1+\frac{1}{n^2}\right)}{\frac{1}{n}}

    use L'Hopital for the indeterminate form \frac{0}{0} ...

    \ln{y} = \lim_{n \to \infty} \frac{-\frac{2}{n^3}}{1 + \frac{1}{n^2}}} \cdot -\frac{n^2}{1}}

    \ln{y} = \lim_{n \to \infty} \frac{\frac{2}{n}}{1 + \frac{1}{n^2}}} = 0

    y = e^0 = 1
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    Re: Converging sequence

    Thank you!
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