limn→∞1en√n∞∑k=0nkk!|n−k|=√2/π
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...
Answer
This question has a nice probabilistic interpretation. Given that X is a Poisson distribution with parameter λ=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∼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:
1en√n+∞∑k=0nkk!|n−k|≈1√n⋅1√2πn∫+∞−∞|x−n|exp(−(x−n)22n)dx=2n√2π∫+∞0xexp(−x22n)dx=√2π,
so the limit is not zero.
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