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Thread: Limit of a series

  1. #1
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    Limit of a series

    Hi,
    Prove that the attached series converges to 1

    Thank's in advance
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  2. #2
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    Re: Limit of a series

    Quote Originally Posted by hedi View Post
    Prove that the attached series converges to 1
    There is a major mistake in your attachment. What is the $\Large k~?$
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Limit of a series

    Quote Originally Posted by hedi View Post
    Hi,
    Prove that the attached series converges to 1

    Thank's in advance
    Here's the sum in question:
    $\displaystyle \sum_{n}^{\infty} \dfrac{(-1)^k}{(k!)^2 {n \choose k}}$

    -Dan
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    Re: Limit of a series

    the series should be
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    Re: Limit of a series

    Quote Originally Posted by hedi View Post
    the series should be
    Now is this correct: $\displaystyle \Large\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {\frac{{{{( - 1)}^k}}}{{{{(k!)}^2}\dbinom{2n}{k}}}}~? $
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  6. #6
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    Re: Limit of a series

    Sorry, it is my mistake.in the denominator there is n,not 2n
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    Re: Limit of a series

    Quote Originally Posted by hedi View Post
    Sorry, it is my mistake.in the denominator there is n,not 2n
    Hedi. Why can't you get this right?
    I have no appetite for spending time on something that you may change on the next post.
    Someone may help you but it will not be me. Sorry.
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  8. #8
    Forum Admin topsquark's Avatar
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    Re: Limit of a series

    So is this (finally) what the question is asking you to calculate?
    $\displaystyle \lim_{n \to \infty} \sum_{k = 0}^m \dfrac{(-1)^k}{(k!)^2 {n \choose k}}$

    Note that $\displaystyle {n \choose k} = \dfrac{n!}{k! ~ (n - k)!}$

    So your problem is
    $\displaystyle \lim_{n \to \infty} \sum_{k = 0}^m \dfrac{(-1)^k}{(k!)^2 {n \choose k}} = \lim_{n \to \infty} \sum_{k = 0}^n \dfrac{(-1)^k}{(k!)^2 \dfrac{n!}{k! ~ (n - k)!}}$

    $\displaystyle = \lim_{n \to \infty} \sum_{k = 0}^n \dfrac{(-1)^k k! ~ (n - k)!}{ (k!)^2 n!}$

    $\displaystyle = \lim_{n \to \infty} \sum_{k = 0}^n \dfrac{(-1)^k (n - k)!}{ k! ~ n!}$

    What's next?

    -Dan
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  9. #9
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    Re: Limit of a series

    Let

    $\displaystyle s_n=\sum _{k=0}^n \frac{(-1)^k(n-k)!}{k!n!}$

    expanding we get

    $\displaystyle s_n=1-\frac{1}{n}+\frac{1}{2!n(n-1)}-\frac{1}{3!n(n-1)(n-2)}+\text{...}$

    the terms are decreasing in absolute value so

    $\displaystyle 1-\frac{1}{n}\leq s_n\leq 1$
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