Prove that $\prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$

Question

This result, $$\prod_{k=1}^{\infty} \big\{\big(1+\frac1{k}\big)^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$$ is in a paper by Hirschhorn in the current issue of the Fibonacci Quarterly (vol. 51, no. 2).

I thought it was quite interesting in that it shows how close $(1+\frac1{k})^{k+\frac1{2}}$ is to $e$, and that it would be an interesting challenge problem. The proof there is not too difficult.

This result is a lemma in the proof of the following, which is the principal result in the paper:

$$\prod_{k=0}^n \binom{n}{k} \sim C^{-1}\frac{e^{n(n+2)/2}}{n^{(3n+2)/6}(2\pi)^{(2n+1)/4}} \exp\big\{-\sum_{p\ge 1}\frac{B_{p+1}+B_{p+2}}{p(p+1)}\frac1{n^p}\big\}\text{ as }n \to \infty $$ where $$\begin{align} C &= \lim_{n \to \infty} \frac1{n^{1/12}} \prod_{k=1}^n \big\{k!\big/\sqrt{2\pi k}\big(\frac{k}{e}\big)^k\big\}\\ &\approx 1.04633506677...\\ \end{align} $$ and the $\{B_p\}$ are the Bernoulli numbers, defined by $$\sum_{p \ge 0} B_p\frac{x^p}{p!} = \frac{x}{e^x-1} .$$

I don't expect anyone here to prove this, since Hirschhorn takes over seven pages of involved math to prove it.

Also, it was an interesting exercise in $\LaTeX$ to enter these formulae so that they displayed exactly (or, at least, pretty closely) as in Hirschhorn's article. Among other things, I learned (after a little searching) that a tilde (~) is entered as "\tilde{}".

Answer 1

Let $$S_N = -N +\log \left(\prod_{k=1}^N \left(1+\dfrac1k\right)^{k+1/2}\right)$$ What we want is $\lim_{N \to \infty} \exp\left(S_N\right)$.

We now have \begin{align} S_N & = -N + \sum_{k=1}^N \left(k+\dfrac12\right) \left(\log(k+1) - \log(k)\right)\\ & = - N+ \sum_{k=1}^N (k+1) \log(k+1) - \sum_{k=1}^N k \log(k) - \sum_{k=1}^N \dfrac{\log(k+1) + \log(k)}2\\ & = - N + (N+1) \log(N+1) - \dfrac{\log(N+1)}2 - \log(N!) \end{align} Hence, we get that $$S_N = - N + \left(N+\dfrac12\right) \log(N+1) - \log(N!)$$ From Stirling, we have $$\log(N!) = N \log N - N + \dfrac12 \log(2 \pi N) + \mathcal{O}(1/N)$$ This gives us $$\log(N!) + N = N \log(N) + \dfrac{\log(2\pi)}2 + \dfrac{\log(N)}2 + \mathcal{O}(1/N)$$ Hence, we have $$S_N = \left(N+\dfrac12\right) \log \left(1 + \dfrac1N\right) - \dfrac{\log(2 \pi )}2 + \mathcal{O}(1/N)$$ Now $$\lim_{N \to \infty} S_N = 1 - \dfrac{\log(2 \pi )}2$$ Hence, the answer you want is $$\exp\left(1 - \dfrac{\log(2 \pi )}2\right) = \dfrac{e}{\sqrt{2\pi}}$$

Answer 2

After some algebra, the product becomes

$$e \left( \lim_{k\rightarrow \infty} \frac{(k+1)!}{\sqrt{k+1}((k+1)/e)^{k+1}} \right)^{-1} .$$

The expression in the parenthesis is $\sqrt{2\pi}$ using Stirling's approximation.

Answer 3

Let me put the things into what I believe to be the right context for this type of infinite products.

• The proposed product appears as a part of the-so called Barnes $G$-function which is mainly characterized by the recursion relation $G(z+1)=\Gamma(z) G(z)$ and normalization $G(1)=1$. For positive integer $z$ this defines a kind of superfactorial: $$G(n)=\prod_{k=0}^{n-2}k!\tag{1}$$

• This function has the following infinite product representation: $$G(1+z)=\left(2\pi\right)^{\frac{z}{2}}\exp\left(-\frac{z+z^2\left(1+\gamma\right)}{2}\right)\prod_{k=1}^{\infty} \left(1+\frac{z}{k}\right)^k \exp\left(\frac{z^2}{2k}-z\right),\tag{2}$$ where $\gamma=-\psi(1)$ is the Euler-Mascheroni constant.

• Setting $z=1$ in (2) and using (1), we obtain $$1=\sqrt{2\pi}e^{-1-\gamma/2}\prod_{k=1}^{\infty} \left(1+\frac{1}{k}\right)^k \exp\left(\frac{1}{2k}-1\right).\tag{3}$$

The formula (3) means that the required identity is equivalent to $$\sum_{k=1}^{\infty}\left[\frac{1}{2k}-\frac{1}{2}\ln\left(1+\frac{1}{k}\right)\right]=\frac{\gamma}{2}.$$ But this is almost the definition of $\gamma$ (note that the sum of logarithms is telescoping).