40
$\begingroup$

I am interested about some infinite product representations of $\pi$ and $e$ like this.
Last week I found this formula on internet

$$\pi=e^{3/2}\prod_{n=2}^{\infty}e\left(1-\frac{1}{n^2}\right)^{n^2}$$

which looks like unbelievable.
(I forgot the link but I am sure that this is the formula.)
How can I start to prove this formula?
Thank You.

$\endgroup$
  • 3
    $\begingroup$ There's an $e$ involved and an $n^2$ for some reason that alone convinces me :/ $\endgroup$ – Alec Teal Jul 2 '15 at 8:51
  • 3
    $\begingroup$ It is equivalent to proving $$\sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1} = \frac32 - \ln\pi$$ $\endgroup$ – nospoon Jul 2 '15 at 8:55
  • 7
    $\begingroup$ @corindo: $e^{\frac52}\not= e^{\frac32}\prod e.$ $\endgroup$ – Bumblebee Jul 2 '15 at 8:56
  • 1
    $\begingroup$ Hint: take log to make it into a sum. Factorize $1-1/n^2=(n-1)(n+1)/n^2$ and expand the log. Try to construct telescoping series. Finish using sterlings. $\endgroup$ – Winther Jul 2 '15 at 9:28
  • 1
    $\begingroup$ @nospoon: in this answer, I compute just that sum :-) $\endgroup$ – robjohn Jul 10 '15 at 14:43
55
$\begingroup$

You may write, for $N \geq 2$, $$ \begin{align} e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2}&=e^{3/2}\times\prod_{n=2}^{N}e\times\prod_{n=2}^{N}\left(1-\dfrac{1}{n^2}\right)^{n^2}\\\\ &=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{n^2}}\dfrac{(n+1)^{n^2}}{n^{n^2}}\\\\ &=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{(n+1)^2}}\times\dfrac{(n+1)^{n^2}}{n^{(n-1)^2}}\times n^2\\\\ &=e^{N+1/2}\times\color{blue}{\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{(n+1)^2}}}\times\color{#C00000}{\prod_{n=2}^{N}\dfrac{(n+1)^{n^2}}{n^{(n-1)^2}}}\times \color{green}{\prod_{n=2}^{N}n^2}\\\\ &=e^{N+1/2}\times\color{blue}{\dfrac{1}{N^{(N+1)^2}}}\times\color{#C00000}{\dfrac{(N+1)^{N^2}}{2}}\times \color{green}{ (N!)^2}\\\\ &=\frac12\times e^{N+1/2}\times \left(1+\frac1N\right)^{N^2}\times\dfrac{ (N!)^2}{N^{2N+1}}. \tag1 \end{align} $$ Then one may observe that, as $N \to +\infty$, $$ N^2 \ln \left(1+\frac1N \right)=N-\frac{1}{2}+O\left(\frac1N\right) $$ gives $$ e^{N+1/2}\times\left(1+\frac1N\right)^{N^2}=e^{2N}\left(1+O\left(\frac1N\right)\right) \tag2 $$ and from the Stirling formula, we get $$ \begin{align} (N!)^2&=2\pi \;N^{2N+1}e^{-2N}\left(1+O\left(\frac1N\right)\right) \end{align} $$ $$ \begin{align} \frac{(N!)^2}{N^{2N+1}}&=2\pi \;e^{-2N}\left(1+O\left(\frac1N\right)\right).\tag3 \end{align} $$ By combining $(1)$, $(2)$ and $(3)$ we obtain, as $N \to +\infty$,

$$ e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2} = {\large \pi} \left(1+O\left(\frac1N\right)\right) $$

leading to the desired result.

$\endgroup$
  • 1
    $\begingroup$ How...Just... Okay. Cool. $\endgroup$ – Kugelblitz Jul 2 '15 at 10:15
  • 5
    $\begingroup$ very simple and beautiful answer +1 $\endgroup$ – Paramanand Singh Jul 2 '15 at 11:14
  • 2
    $\begingroup$ What an amazing solution! (+1) The log approximation was a masterstroke. $\endgroup$ – hypergeometric Aug 8 '15 at 17:56
16
$\begingroup$

One can write the log of the product as

$$\sum_{n=2}^{\infty} \left [1+n^2 \log{\left (1-\frac1{n^2} \right )} \right ] $$

Now,

$$\log{\left (1-\frac1{n^2} \right )} = -\int_0^1 \frac{du}{n^2-u} $$

So the sum is equal to

$$-\int_0^1 du \, u \sum_{n=2}^{\infty} \frac1{n^2-u} $$

$$\sum_{n=-\infty}^{\infty} \frac1{n^2-u} = -\frac{\pi \cot{\pi \sqrt{u}}}{\sqrt{u}} $$

We get that the sum in question may be written as an integral:

$$\frac12 \int_0^1 du \left (\pi \sqrt{u} \cot{\pi \sqrt{u}} + \frac{2 u}{1-u} - 1 \right )$$

Manipulate a bit and sub $u=v^2$ to get

$$\int_0^1 dv \left (\pi v^2 \cot{\pi v} + \frac{2 v}{1-v^2} \right ) - \frac32 $$

We evaluate the integral by simply evaluating the antiderivative and using the Fundamental Theorem. Using the fact that

$$\cot{\pi v} = 2 \operatorname{Im}{\left (\frac1{1-e^{-i 2 \pi v}} \right )}$$

and integrating by parts, we find that

$$\pi \int dv \, v^2 \cot{\pi v} = \frac{i v \text{Li}_2\left(e^{-2 i \pi v}\right)}{\pi}+\frac{\text{Li}_3\left(e^{-2 i \pi v}\right)}{2 \pi^2}+\frac{i \pi v^3}{3}+v^2 \log \left(1-e^{-2 i \pi v}\right) + C$$

$$\int dv \, \frac{2 v}{1-v^2} = -\log{(1-v^2)} + C$$

In taking the integral from $v=0$ to $v=1$, the $\text{Li}_3$ term vanishes. The other terms vanish at $v=0$, so we need only concern ourselves with the limit of the sum of the expressions as $v \to 1^-$. We get, as the limit,

$$\frac{i \pi}{6} + \frac{i \pi}{3} + \log{(-i 2 \pi)} - \log{2} = \log{\pi} $$

(NB $\text{Li}_2(1) = \pi^2/6$.) The log of the product is therefore

$$\log{\pi} - \frac32$$

From this, exponentiating produces the original result.

$\endgroup$
  • $\begingroup$ +1 I'm following the same route, get similar expression but never able to complete the transform. $\endgroup$ – achille hui Jul 2 '15 at 9:55
  • $\begingroup$ @achillehui: I think it is a simple as evaluating the antiderivative and taking the limit near the singular point at $v=1$. $\endgroup$ – Ron Gordon Jul 2 '15 at 9:56
  • $\begingroup$ +1, I was studying the same integral using the closed form of $\sum_{n\geq1}\left(\zeta\left(2n\right)-1\right)x^{2n}.$ Nice! $\endgroup$ – Marco Cantarini Jul 2 '15 at 10:26
  • $\begingroup$ A minor thing, but in line 2 $$ - \int_{0}^{1} \frac{du}{n^{2} - u} = \log \bigg( 1 - \frac{1}{n^{2}} \bigg) \ne n^2 \log \bigg( 1 - \frac{1}{n^{2}} \bigg)$$ $\endgroup$ – Mattos Jul 2 '15 at 13:46
  • 1
    $\begingroup$ Similarly, we have $~\displaystyle\prod_{n=1}^\infty \bigg[~e~\bigg(1+\frac1{n^2}\bigg)^{-n^2}~\bigg] ~=~ \dfrac1{\displaystyle\prod_{n=1}^\infty \bigg[~\frac1e~\bigg(1+\frac1{n^2}\bigg)^{n^2}~\bigg]} ~=~ e^S~,~$ where $$S~=~\ln\Big(e^{2\pi}-1\Big) ~-~ \bigg(\frac\pi3+\frac12\bigg) ~+~ \frac1\pi\cdot\Re\bigg[\text{Li}_2\Big(e^{2\pi}\Big) ~+~ \frac{\zeta(3)-\text{Li}_3\Big(e^{2\pi}\Big)}{2\pi}\bigg].$$ $\endgroup$ – Lucian Jul 9 '15 at 23:20
11
$\begingroup$

Taking logs and using the power series for $\log(1-x)$, we get $$ \begin{align} \frac32+\sum_{n=2}^\infty\left[1+n^2\log\left(1-\frac1{n^2}\right)\right] &=\frac32-\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{(k+1)n^{2k}}\\ &=\frac32-\sum_{k=1}^\infty\frac{\zeta(2k)-1}{k+1}\tag{1} \end{align} $$ Transcribing $(5)$ from this answer: $$ \begin{align} &\sum_{n=1}^\infty\frac{\zeta(2n)-1}{n+1}\\ &=\sum_{n=1}^\infty\int_0^1(\zeta(2n)-1)\,2x^{2n+1}\,\mathrm{d}x\tag{2a}\\ &=\int_0^1\left(1-\pi x\cot(\pi x)-\frac{2x^2}{1-x^2}\right)\,x\,\mathrm{d}x\tag{2b} \\ &=\int_0^1\left(3-\pi\cot(\pi x)-\frac2{1-x^2}\right)\,x\,\mathrm{d}x\tag{2c}\\ &=\frac32-\lim_{\lambda\to1^-}\int_0^\lambda\left(\pi\cot(\pi x)+\frac2{1-x^2}\right)\,x\,\mathrm{d}x\tag{2d}\\ &=\frac32-\lim_{\lambda\to1^-}\left[\vphantom{\sum}\lambda\log(\sin(\pi\lambda))-\log(1-\lambda^2)\right]+\int_0^1\log(\sin(\pi x))\,\mathrm{d}x\tag{2e}\\ &=\frac32-\log(\pi)+\log(2)-\log(2)\tag{2f}\\ &=\frac32-\log(\pi)\tag{2g} \end{align} $$ Explanation:
$\text{(2a)}$: $\int_0^12x^{2n+1}\,\mathrm{d}x=\frac1{n+1}$
$\text{(2b)}$: use the generating function for $\zeta(2n)$ derived in this answer
$\text{(2c)}$: use that $1-\frac{2x^2}{1-x^2}=3-\frac2{1-x^2}$ and $(4)$
$\text{(2d)}$: write the integral as a limit
$\text{(2e)}$: integrate by parts
$\text{(2f)}$: use $\lim\limits_{\lambda\to1}\frac{\sin(\pi \lambda)}{1-\lambda}=\pi$ and $(6)$
$\text{(2g)}$: cancel $\log(2)$

Combining $(1)$ and $(2)$, we get $$ \bbox[5px,border:2px solid #C0A000]{e^{3/2}\prod_{n=2}^\infty e\left(1-\frac1{n^2}\right)^{n^2}=\pi}\tag{3} $$


Results Used in $\boldsymbol{(2)}$

The substitution $x\mapsto1-x$ shows that the following integral equals its negative. Therefore, $$ \int_0^1\pi(1-x)\cot(\pi x)\,x\,\mathrm{d}x=0\tag{4} $$ Substituting $x\mapsto2x$ yields $$ \begin{align} &\int_0^1\log(\sin(\pi x))\,\mathrm{d}x\\ &=2\int_0^{1/2}\log(\sin(2\pi x))\,\mathrm{d}x\\ &=\log(2)+2\int_0^{1/2}\log(\sin(\pi x))\,\mathrm{d}x+2\int_0^{1/2}\log(\cos(\pi x))\,\mathrm{d}x\\ &=\log(2)+2\int_0^{1/2}\log(\sin(\pi x))\,\mathrm{d}x+2\int_{1/2}^1\log(\sin(\pi x))\,\mathrm{d}x\\ &=\log(2)+2\int_0^1\log(\sin(\pi x))\,\mathrm{d}x\tag{5} \end{align} $$ Therefore, $$ \int_0^1\log(\sin(\pi x))\,\mathrm{d}x=-\log(2)\tag{6} $$

$\endgroup$
  • $\begingroup$ Nice answer @robjohn. Thank you. +1. $\endgroup$ – Bumblebee Jul 13 '15 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.