$n$th Mathematics Blog

A place where I write about math-related topics

Product of an Infinite Series

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Recently, I came across a problem that was asking for a way to calculate the infinite product of $1\cdot 2\cdot 3\cdot 4\cdot\ldots$. At first glance I thought there wasn’t an answer to this since the sequence diverges. However, there is a way to calculate this product and in this post I will show just that. However, it is arguable whether this is really calculating the product.

First note that

$$\prod_{n=1}^{\infty}n=\exp\left[\log\left(\prod_{n=1}^{\infty}n\right)\right]=\exp\left(\sum_{n=1}^{\infty}\log{n}\right)$$ Now, notice that $$\zeta’(s)=\frac{d}{ds}\sum_{n=1}^{\infty}n^{-s}=\sum_{n=1}^{\infty}\frac{d}{ds}n^{-s}=\sum_{n=1}^{\infty}-n^{-s}\log{n}=-\sum_{n=1}^{\infty}n^{-s}\log{n}$$ At $s=0$, $$\zeta’(0)=-\sum_{n=1}^{\infty}\log{n}$$ $$\therefore \exp\left(\sum_{n=1}^{\infty}\log{n}\right)=\exp(-\zeta’(0))$$

Now what is $\sum_{n=1}^{\infty}\log{n}$?

Note that by Wallis’s formula, we have that: $$\frac{\pi}{2}=\prod_{n=1}^{\infty}\frac{(2n)^2}{(2n-1)(2n+1)}=\frac{2\cdot 2}{1\cdot 3}\cdot\frac{4\cdot 4}{3\cdot 5}\cdot\ldots$$ Now, consider this function: $$F(s)=\left(1-2^{1-s}\right)\zeta(s)=\frac 12 + \frac 12\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)$$ $$\therefore F’(s)=2^{1-s}(\log{2})\zeta(s)+\left(1-2^{1-s}\right)\zeta’(s)=\frac 12\sum_{n=1}^{\infty}(-1)^n\left[\frac{\log{n}}{n^s}-\frac{\log(n+1)}{(n+1)^s}\right]$$ Note that $\zeta(0)=-\frac{1}{2}$, then at $s=0$: $$F’(0)=2(\log{2})\left(-\frac{1}{2}\right)+(1-2)\zeta’(0)=\frac 12\sum_{n=1}^{\infty}(-1)^n[\log{n}-\log(n+1)]$$ $$=\frac{1}{2}\log\left(\frac{2\cdot 2}{1\cdot 3}\cdot\frac{4\cdot 4}{3\cdot 5}\cdot\ldots\right)=\frac{1}{2}\log{\frac{\pi}{2}}$$ $$\therefore \zeta’(0)=-\frac{1}{2}\log{2\pi}$$

$$\therefore \prod_{n=1}^{\infty}n=\exp\left(\sum_{n=1}^{\infty}\log{n}\right)=\exp(-\zeta’(0))=\sqrt{2\pi}$$

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