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I was reading a proof on the evaluation of $\int_0^\infty e^{-x^2}\ dx$ without advanced techniques and stumbled upon two limits that I can't seem to crack: $$\lim_{m\to\infty}\left(\sqrt{m}\cdot\prod_{n=1}^m\frac{2n}{2n+1}\right)=\frac{\sqrt{\pi}}2$$ $$\lim_{m\to\infty}\left(\sqrt{m}\cdot\prod_{n=2}^m\frac{2n-3}{2n-2}\right)=\frac1{\sqrt{\pi}}$$ The proof does not go into detail on how these limits were obtained, and since I wanted to understand it completely, I thought this would be the best place to ask. I have not been exposed to infinite products (only summations) and therefore I do not know which rules to apply (I feel as if they are quite similar?). In both cases, I see that an indeterminate form $0\cdot\infty$ presents its self, therefore I am guessing Hospital would be a nice approach? Any help is appreciated! Also, my calculus book does not tackle infinite products, any suggestions on books that might give me a general outlook on the subject?

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    $\begingroup$ Wikipedia has an article on Wallis's integrals, which are probably the easiest way to show these: it was the way used in a first-year university course in probability. See in particular the section "Equivalence". $\endgroup$ – Chappers Mar 16 '17 at 13:44
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Wallis's formula:

$$\frac{\pi}{2}=\prod_{n=1}^\infty \left[\frac{(2n)^2}{(2n+1)(2n-1)}\right].$$

Proof: Weierstrass factorization of $\sin$ (You can find Euler's semi standard proof of this here) : $$\sin(x)=x\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right).$$

Plug in $x=\pi/2$ and play with the resulting fractions to get the desired result.

For your first product:

\begin{align*} \prod_{n=1}^m\frac{2n}{2n+1}&=\frac{2\cdot 1}{2\cdot 1+1}\frac{2\cdot 2}{2\cdot 2+1}\frac{2\cdot 3}{2\cdot 3+1}\cdots \frac{2\cdot m}{2\cdot m+1}\\ &=2\cdot 1\frac{2\cdot 2}{2\cdot 1+1}\frac{2\cdot 3}{2\cdot 2+1}\cdots \frac{2\cdot m}{2\cdot (m-1)+1}\frac{1}{2m+1}\\ &=\frac{2}{2m+1}\prod_{n=2}^m\frac{2n}{2n-1} \end{align*}

Thus:

\begin{align*} \frac{\pi}{2}&=\lim_{m\rightarrow\infty}\prod_{n=1}^m \left[\frac{(2n)^2}{(2n+1)(2n-1)}\right]\\ &=\lim_{m\rightarrow\infty}\left(\prod_{n=1}^m\frac{2n}{2n+1}\right)\frac{1}{2}\left(\prod_{n=2}^m\frac{2n}{2n-1}\right)\\ &=\lim_{m\rightarrow\infty}\frac{1}{2}\frac{2m+1}{2}\left(\prod_{n=1}^m\frac{2n}{2n+1}\right)^2. \end{align*}

Now just take the square-root of both sides and notice that $\sqrt{m}/\sqrt{\frac{2m+1}{2}}\rightarrow 1$

For the second question, try a similar trick by shifting the index $n\rightarrow n+2$.

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  • $\begingroup$ I can't seem to grasp how you obtained $\lim_{m\rightarrow\infty}\frac{2m+1}{2}\left(\prod_{n=1}^m\frac{2n}{2n+1}\right)^2$ from the first relation... shouldn't it be $\lim_{m\rightarrow\infty}\frac2{2m+1}\left(\prod_{n=1}^m\frac{2n}{2n-1}\right)^2$? $\endgroup$ – user372003 Mar 16 '17 at 5:09
  • $\begingroup$ Also, where would $\sqrt{m}$ come from? Thanks! $\endgroup$ – user372003 Mar 16 '17 at 5:15
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    $\begingroup$ @Denis: Throw the $2/(2m+1)$ to the other side of the first product. In other words, either product can be written in terms of the other. For the $\sqrt{m}$, just write: $\frac{2m+1}{2}=m\cdot \frac{2m+1}{2m}$ and in the limit the second fraction will become 1. To get $\sqrt{m}$, take the squareroot of both sides and pass it through the limit. $\endgroup$ – Alex R. Mar 16 '17 at 18:19
  • $\begingroup$ Thank you very much very clear, however when I do the calculation I get $\sqrt{\frac\pi2}$ and not $\frac{\sqrt\pi}2$. $\endgroup$ – user372003 Mar 16 '17 at 19:11
  • $\begingroup$ @Denis: Good catch! I lost a factor of 2. the product for $\prod_{n=1}^m\frac{2n}{2n-1}$ starts at $n=1$, whereas my formulation was from $n=2$. I've made the correct above. $\endgroup$ – Alex R. Mar 16 '17 at 19:23
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Both of these can be obtained as a consequence of Stirling's approximation, by first rewriting all of the partial products in terms of factorials. This argument doesn't seem easier to me than the standard argument involving passing to a 2-dimensional integral.

In general, a standard strategy for handling infinite products is to take their logarithms, producing infinite sums.

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  • $\begingroup$ I haven't been formerly introduced to double integration, since I am currently taking Calculus II. However, I am going to take a look at the link you posted. $\endgroup$ – user372003 Mar 16 '17 at 4:28
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{m \to \infty}\pars{\root{m}\prod_{n = 1}^{m}{2n \over 2 n + 1}} = \lim_{m \to \infty}{\root{m}\,m! \over \prod_{n = 1}^{m}\pars{n + 1/2}} = \lim_{m \to \infty}{\root{m}\,m! \over \pars{3/2}^{\overline{m}}} \\[5mm] = &\ \lim_{m \to \infty}{\root{m}\,m! \over \Gamma\pars{3/2 + m}/\Gamma\pars{3/2}} = \Gamma\pars{3 \over 2}\lim_{m \to \infty} {m^{1/2}\root{2\pi}m^{m + 1/2}\expo{-m} \over \root{2\pi}\pars{m + 1/2}^{m + 1}\expo{-\pars{m + 1/2}}} \\[5mm] = &\ {1 \over 2}\,\root{\pi}\lim_{m \to \infty}{\expo{1/2} \over \bracks{1 + \pars{1/2}/m}^{\,m + 1}} = \bbx{\ds{\root{\pi} \over 2}} \end{align}

The other one follows the same pattern.

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