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Math Help - A series limit

  1. #1
    Moo
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    A series limit

    Hi !

    Prove that :

    \lim_{n \to \infty} ~ e^{-n} \sum_{k=0}^n \frac{n^k}{k!}=\frac 12

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  2. #2
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    Quote Originally Posted by Moo View Post
    Hi !

    Prove that :

    \lim_{n \to \infty} ~ e^{-n} \sum_{k=0}^n \frac{n^k}{k!}=\frac 12


    i think the limit originally comes from probability, something that i'm a complete idiot at it! there's the idea of the proof in this thread. i know a calculus proof but it's long and very ugly!

    so i think it's a good challenge finding a neat calculus (analysis) solution for this question.
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  3. #3
    Moo
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    Quote Originally Posted by NonCommAlg View Post
    i think the limit originally comes from probability, something that i'm a complete idiot at it! there's the idea of the proof in this thread. i know a calculus proof but it's long and very ugly!

    so i think it's a good challenge finding a neat calculus (analysis) solution for this question.
    Yes, that was using a probability proof
    Same type as the previous ones :P


    Laurent in the thread gave the way to solve it, I'll detail it here


    Consider a sequence (X_n)_{n \geq 1} of independent random variables, following a Poisson distribution with parameter 1 :
    \mathbb{P}(X_i=k)=e^{-1} \cdot \frac{1}{k!}

    Then let's consider S_n=X_1+\dots+X_n
    It is easy to prove that it follows a Poisson distribution with parameter n :
    \mathbb{P}(S_n=k)=e^{-n} \cdot \frac{n^k}{k!}

    We can see that :
    \mathbb{P}(S_n \leq n)=\sum_{k=0}^n e^{-n} \cdot \frac{n^k}{k!}
    But \{S_n \leq n\}=\{S_n-n\leq 0\}=\left\{\frac{S_n-n}{\sqrt{n}} \leq 0\right\}

    Hence \mathbb{P}\left(\frac{S_n-n}{\sqrt{n}} \leq 0\right)=e^{-n} \sum_{k=0}^n \frac{n^k}{k!}


    Then we must check the conditions for applying the Central Limit Theorem :
    \mathbb{V}\text{ar}(X_i)=\mathbb{E}(X_i)=1<\infty \Rightarrow X_i \in L^2
    And the X_i are independent and identically distributed.
    The Central Limit Theorem says that :
    \frac{S_n-n \mathbb{E}(X_i)}{\sqrt{n}}=\frac{S_n-n}{\sqrt{n}} converges to the standard normal distribution \mathcal{N}(0,1)

    In particular, there is a convergence of the cumulative density functions :
    \mathbb{P}\left(\frac{S_n-n}{\sqrt{n}} \leq 0\right) \longrightarrow \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} e^{-t^2/2} ~dt, which is \frac 12 because the integrand is even.

    This finishes the proof :
    e^{-n} \sum_{k=0}^n \frac{n^k}{k!} \longrightarrow \frac 12

    Haaa... I really like these kinds of limits that use probability, it's just great



    Now, if you have time, I'd be interested in seeing this proof you have that uses calculus ^^
    Last edited by Moo; December 20th 2011 at 01:40 PM. Reason: correcting the LaTeX
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    sure, if i basically can find the solution i had because it goes back to quite long time ago when i was a calculus I student! haha
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