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Thread: Sequence Problem with Parenthesis

  1. #1
    MHF Contributor Jason76's Avatar
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    Sequence Problem with Parenthesis

    a_{n} = (\dfrac{n+1}{2n})(1 - \dfrac{1}{n})

    Convergent or divergent?

    \lim n\to \infty[(\dfrac{n+1}{2n})(1 - \dfrac{1}{n})]

    [(\dfrac{\infty+1}{2\infty})(1 - \dfrac{1}{\infty})]

    [(indeterminate) (1)] = indeterminate

    Lost on this one. I know the book says the answer is (convergent to 1/2). Do we expand this expression before taking a limit?
    Last edited by Jason76; Nov 10th 2014 at 07:04 PM.
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  2. #2
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    Prove It's Avatar
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    Re: Sequence Problem with Parenthesis

    No, the limit of a product is equal to the product of the limits.

    Notice $\displaystyle \begin{align*} \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n} \to \frac{1}{2} \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$.

    Also $\displaystyle \begin{align*} 1 - \frac{1}{n} \to 1 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$

    Thus $\displaystyle \begin{align*} \left( \frac{n + 1}{2n} \right) \left( 1 - \frac{1}{n} \right) \to \frac{1}{2} \cdot 1 = \frac{1}{2} \end{align*}$ as $\displaystyle \begin{align*}n \to \infty \end{align*}$.
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  3. #3
    MHF Contributor Jason76's Avatar
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    Re: Sequence Problem with Parenthesis

    Quote Originally Posted by Prove It View Post
    No, the limit of a product is equal to the product of the limits.

    Notice $\displaystyle \begin{align*} \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n} \to \frac{1}{2} \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$.

    Also $\displaystyle \begin{align*} 1 - \frac{1}{n} \to 1 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$

    Thus $\displaystyle \begin{align*} \left( \frac{n + 1}{2n} \right) \left( 1 - \frac{1}{n} \right) \to \frac{1}{2} \cdot 1 = \frac{1}{2} \end{align*}$ as $\displaystyle \begin{align*}n \to \infty \end{align*}$.
    Ok sounds great. I get it.
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