# How to evaluate the following limits?

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?

• 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". – Chappers Mar 16 '17 at 13:44

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.

\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 $n\rightarrow n+2$.

• 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 $\lim_{m\rightarrow\infty}\frac2{2m+1}\left(\prod_{n=1}^m\frac{2n}{2n-1}\right)^2$? – user372003 Mar 16 '17 at 5:09
• Also, where would $\sqrt{m}$ come from? Thanks! – user372003 Mar 16 '17 at 5:15
• @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. – Alex R. Mar 16 '17 at 18:19
• Thank you very much very clear, however when I do the calculation I get $\sqrt{\frac\pi2}$ and not $\frac{\sqrt\pi}2$. – user372003 Mar 16 '17 at 19:11
• @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. – Alex R. Mar 16 '17 at 19:23

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.

• 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. – user372003 Mar 16 '17 at 4:28


The other one follows the same pattern.