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I hope it's okay to ask this, but I am trying to understand the following solution done by Felix Marin to the well-known limit: $$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}=\frac12$$ I can't understand the following line (and from the comments I think I am not the only one): $$\lim_{n\to \infty}\left(e^{-n}\sum_{k=0}^n\exp{(k\ln n-\ln(k!))}\right)=\lim_{n\to \infty}\left(e^{-n}\sum_{k=0}^n\exp{(n\ln n-\ln(n!))-\frac1{2n}(k-n)^2}\right)$$

So I'll try to keep the question simple, can anyone please explain in more details that equality?

Or alternatively to prove in a different way that: $$\lim_{n\to \infty}\left(e^{-n}\sum_{k=0}^n\exp{(k\ln n-\ln(k!))}\right)=\lim_{n\to\infty}\left(\frac{e^{-n} n^n}{n!}\sum_{k=0}^n\exp\left(-\frac{1}{2n}(k-n)^2\right)\right)$$

PinkyWay
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Zacky
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    I assume the usual probabilistic proof is not satisfying here? – Ian Apr 27 '19 at 19:13
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    Related – mrtaurho Apr 27 '19 at 19:13
  • Indeed, I am trying to understand this way, and I think it's quite smooth beside that equality. – Zacky Apr 27 '19 at 19:13
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    It's not straightforward to actually do everything rigorously within Felix's approach. But basically the idea of that line is to obtain the quadratic Taylor approximation of $x \ln(n) - \ln(\Gamma(x+1))-n$ (where I have brought the $e^{-n}$ inside the sum) in a vicinity of $x=n$, and then hope that the Taylor error somehow washes out in the limit. – Ian Apr 27 '19 at 19:24
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    I think some of the other answers, such as the one by user26872, are along the lines of what you're trying to do here but they rely on a firmer starting point. Namely, they use the explicit, non-asymptotic connection between the exponential partial sum and the incomplete Gamma function. – Ian Apr 27 '19 at 19:32

1 Answers1

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Focusing only on this particular line, we bring the $e^{-n}$ inside and write

$$k \ln(n) - \ln(k!) - n \\= n \ln(n) - n - \ln(n!) + (k-n) \ln(n) + \ln(n!) - \ln(k!).$$

The first three terms by Stirling's approximation are $\ln((2\pi n)^{-1/2})+o(1)$.

To handle $\ln(n!)-\ln(k!)$ we might hope to replace it by $\ln(\Gamma(n+1)) - \ln(\Gamma(k+1))$ where $k$ is now thought of as a continuous quantity, and then obtain the Taylor approximation

$$\ln(n!)-\ln(k!) \approx (n-k) \psi^{(0)}(n+1) - \frac{1}{2} (n-k)^2 \psi^{(1)}(n+1)$$

where $\psi^{(j)}$ are polygamma functions. Now we need to know some asymptotics for the $\psi^{(j)}$; one might hope that these correspond to term-by-term differentiation of Stirling's approximation, but this doesn't come for free from any general theorem.

Anyway, it turns out that the results needed here do work out the way that you would expect by differentiating the Stirling approximation. Specifically, $\psi^{(0)}(n+1)=\ln(n)+O(1/n)$ and $\psi^{(1)}(n+1)=1/n+o(1/n)$ as $n \to \infty$. Thus the term $(n-k)(\psi^{(0)}(n+1)-\ln(n))$ becomes significant only when $|n-k|=\Omega(n)$. Meanwhile, $\psi^{(1)}(n+1)(n-k)^2$ becomes significant when $|n-k|=\Omega(n^{1/2})$. The end result is that by the time the linear term is significant, the quadratic term has already made the overall exponential vanishingly small anyway. In this way we obtain the approximation

$$k\ln(n)-\ln(k!)-n \approx \ln((2\pi n)^{-1/2}) - \frac{(n-k)^2}{2n}.$$

You can recognize the first term as the logarithm of the normalization constant of a Gaussian density and the second term as the logarithm of the exponential part of the Gaussian density.

Formalizing this properly really requires you to first justify passing to an integral in the first place so that you can properly apply Watson's lemma in order to justify using derivatives to begin with. I consider that step to be beyond the scope of this question.

In general I think ironing out Felix's approach is quite a bit more difficult than using the explicit relationship between the partial exponential sum and the incomplete Gamma function, as discussed in other answers to that question.

Ian
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