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$$\lim\limits_{n\to\infty}\frac{1}{e^n\sqrt{n}}\sum\limits_{k=0}^{\infty}\frac{n^k}{k!}|n-k|=\sqrt{2/\pi}$$ Is this limit true? I should show limit is true. It is allowed to use computer programs to find this limit. Thanks for your helps...

guest
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2 Answers2

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Split the series at $k = n$:

\begin{align} \sum_{k=0}^\infty \frac{n^k}{k!}\lvert n-k\rvert &= \sum_{k=0}^n \frac{n^k}{k!}(n-k) + \sum_{k=n+1}^\infty \frac{n^k}{k!}(k-n)\\ &= \sum_{k=0}^n\frac{n^{k+1}}{k!} - \sum_{k=1}^n \frac{n^k}{(k-1)!} + \sum_{k=n+1}^\infty \frac{n^k}{(k-1)!} - \sum_{k=n+1}^\infty \frac{n^{k+1}}{k!}\\ &= \sum_{k=0}^n \frac{n^{k+1}}{k!} - \sum_{k=0}^{n-1} \frac{n^{k+1}}{k!} + \sum_{k=n}^\infty \frac{n^{k+1}}{k!} - \sum_{k=n+1}^\infty \frac{n^{k+1}}{k!}\\ &= 2\frac{n^{n+1}}{n!}. \end{align}

So you are looking for

$$\lim_{n\to\infty} \frac{2n^{n+\frac{1}{2}}}{n!e^n}.$$

Now recall or look up Stirling's formula.

Daniel Fischer
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This question has a nice probabilistic interpretation. Given that $X$ is a Poisson distribution with parameter $\lambda=n$, we are essentially computing the expected value of the absolute difference between $X$ and its mean $n$. The central limit theorem gives that $Y\sim N(n,n)$ (a normal distribution with mean and variance equal to $n$) is an excellent approximation of our distribution for large values of $n$, hence: $$\begin{eqnarray*}\frac{1}{e^n \sqrt{n}}\sum_{k=0}^{+\infty}\frac{n^k}{k!}|n-k|&\approx&\frac{1}{\sqrt{n}}\cdot\frac{1}{\sqrt{2\pi n}}\int_{-\infty}^{+\infty}|x-n|\exp\left(-\frac{(x-n)^2}{2n}\right)\,dx\\&=&\frac{2}{n\sqrt{2\pi}}\int_{0}^{+\infty}x\exp\left(-\frac{x^2}{2n}\right)\,dx\\&=&\color{red}{\sqrt{\frac{2}{\pi}}},\end{eqnarray*}$$ so the limit is not zero.

Jack D'Aurizio
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  • Thanks for your help, and I will use it in my study. Any problem? – guest May 11 '15 at 15:04
  • @guest: not at all, you are welcome. – Jack D'Aurizio May 11 '15 at 15:06
  • Pedantry, but I think it might be more usual to write $Y \sim N(n,n)$, i.e. in terms of the mean and variance $N(\mu,\sigma^2)$. This does not affect the rest of your answer. – Henry May 11 '15 at 15:14
  • @Henry: yes, there is an ambiguity in defining the standard parameters for a normal distribution, both $N(\mu,\sigma)$ and $N(\mu,\sigma^2)$ are used. Anyway, I wrote down the pdf. – Jack D'Aurizio May 11 '15 at 15:21
  • I have no knowledge about distributions but it will help me in limit calculations. I want to learn more about distributions. By the way, in fact I have another limit to be calculated. – guest May 12 '15 at 14:32
  • It is $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{\infty}|\ln n-\ln k|\left(1-\frac{1}{n}\right)^k=?$ – guest May 12 '15 at 14:35
  • @guest: better to post it in another question. It is quite different from the limit solved here. – Jack D'Aurizio May 12 '15 at 14:38
  • It seems like geometric distribution when $p=\frac{1}{n}$. But I have no idea about how to find expected value in geometric distribution? – guest May 12 '15 at 14:38
  • @guest: there is a geometric series in disguise and the limit can be computed by using a Riemann sums argument. – Jack D'Aurizio May 12 '15 at 14:40
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    ok ı will ask it as another question. – guest May 12 '15 at 14:42
  • I added the question with title "limit of the expression " – guest May 12 '15 at 14:49