# What is the limit of $\frac{\prod\mathrm{Odd}}{\prod\mathrm{Even}}?$ Does $\pi$ show up here (again)?

What is the result of the following limit?

$$\frac{1\times3\times5\times\cdots}{2\times4\times6\times8\times\cdots} = \lim_{n \rightarrow \infty}\prod_{i=1}^{n}\frac{(2i-1)}{2i}$$

If I remember correctly, it is something related to $$\pi$$.

How can I compute it? And moreover, how can I compute the summation of the series?

• I feel like this is just going to $0$. You might actually mean $\frac{\pi}{2}=\frac{2\cdot2\cdot4\cdot4\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5 \cdots }$
– user304329
Commented Mar 11, 2016 at 18:50
• $\displaystyle \frac{1\times3\times5\times7}{2\times4\times6\times8} = \frac{1\times3\times\cdots \times (2n-1)}{2\times4\times\cdots\times(2n)} \vphantom{\frac {}{\displaystyle\int}}$ when $n=4$. One must consider $\displaystyle\lim_{n\to\infty} \frac{1\times3\times\cdots \times (2n-1)}{2\times4\times\cdots\times(2n)}$ and not $\displaystyle \frac{\lim_{n\to\infty} 1\times3\times\cdots\times(2n-1)}{\lim_{n\to\infty}2\times4\times\cdots\times(2n)} \vphantom{\frac {}{\displaystyle\int}}$. $\qquad$ Commented Mar 11, 2016 at 18:54
• You can also shorten the notation with the double factorial n!! Commented Mar 11, 2016 at 18:59
• @Dan : That double factorial notation is annoying because $n!!$ ought to mean $(n!)!$. $\qquad$ Commented Mar 11, 2016 at 19:21
• What vrugtehagel said. You're probably thinking of the Wallis product for $\pi$ Commented Mar 12, 2016 at 2:27

If you take the numerator product up to $2n-1$ and the denominator up to $2n,$ you have exactly $$\frac{(2n)!}{4^n (n!)^2}$$

By Stirling's approximation this is asymptotic to $$\frac{1}{\sqrt {\pi n}}$$ and goes to $0$ slowly. If we were to add one more term to the numerator, that is $2n+1,$ the expression would go to $\infty$ slowly, as a constant times $\sqrt n.$

The only simple way to get a limit is to take the numerator up to $2n-1$ as before, but also multiply the numerator by a single factor of $\sqrt n.$ Then the limit would be $1/ \sqrt \pi$

• @SubhadeepDey, IF we multiply by an extra $\sqrt n,$ we get a nonzero finite limit. The original thing goes to zero slowly. Commented Mar 11, 2016 at 19:12

Here you need to find $\lim A_n$, where $$A_n= \frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}.$$

Now note that, for $n\gt1$, $2n-1\gt n$. So, in the numerator, you get $$\frac{1\times 2\times 3\dots\times n}{2\times4\times 6\times\dots\times 2n}\lt\frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}\\\implies\frac1{2^n}\lt \frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}$$

Now, you use the identity $$\frac n{n+1}\lt\frac {n+1}{n+2}$$ and define a sequence $$B_n=\frac 23\times \frac 45\times\dots \times \frac {2n}{2n+1} .$$

So, now you have $$A_n\lt B_n\\\implies (A_n)^2\lt A_nB_n=\frac 1{2n+1}\\\implies A_n\lt \frac 1{\sqrt{2n+1}}.$$

So, you have $$\frac 1{2^n}\lt A_n\lt \frac 1{\sqrt{2n+1}}\\\implies\lim \frac 1{2^n}\lt\lim A_n\lt\lim \frac 1{\sqrt{2n+1}}.$$

So, by sandwitch theorem, $$\lim A_n=0.$$

• I didn't downvote, but the first half of the proof is unnecessary. For the sandwich theorem, all you need is $0 < A_n < 1/\sqrt{2n+1}$, and $0 < A_n$ is obvious. Commented Mar 11, 2016 at 23:51
• @alephzero, yes, I know. I have just showed that there are several ways to get the lower sequence.
– user249332
Commented Mar 12, 2016 at 7:49
• JUST BRILLIANT... Commented Mar 12, 2016 at 20:06

As has been pointed out, the limit is zero.

However you mention that you remember it's something involving $\pi$. What you're probably thinking of is something like the result

$$\lim_{n \to \infty} \sqrt{n} {1 \cdot 3 \cdots (2n-1) \over 2 \cdot 4 \cdots 2n} = \frac 1{\sqrt{\pi}}.$$

Notice the additional factor of $\sqrt{n}$ in front. This means that your initial product, with $n$ factors on the top and $n$ factors on the bottom, is near $1/{\sqrt{\pi n}}$ when $n$ is larger. For example if $n = 100$ the product is about 0.05634 and $1/\sqrt{100\pi} \approx 0.05642$.

The limit quoted above can be proved using Stirling's formula,

$$\lim_{n \to \infty} {n! \over \sqrt{2\pi n} (n/e)^n} = 1,$$

if you rewrite the product as $(2n)!/(2^n n!)^2$.

To paraphrase Wikipedia: if $a_n$ is a sequence of complex numbers such that $\sum_{n\geq 1} |a_n|^2 < \infty$, then non-zero convergence of $\prod_{n\geq 1}(1+a_n)$ is equivalent to convergence of $\sum_{n\geq 1} a_n$.

Letting $a_n = -1/2n$, we can see that, because the harmonic series diverges to infinity, your product converges to zero.

• Great. Thanks... Commented Jan 2, 2017 at 20:26

If $0\leq a_n<1$ for all $n,$ then $$\prod_{n=1}^{\infty}(1-a_n)=0\iff \sum_{n=1}^{\infty}a_n=\infty.$$ Let $a_n=1/2 n.$ Then $\sum_{n=1}^{\infty}a_n=\infty.$ .

So $\quad \prod_{n=1}^{\infty}\frac {2 n-1}{2 n}=\prod_{n=1}^{\infty}(1-a_n)=0.$

• Proving the first assertion seems harder than the other proofs of the OP's result - unless it's a standard theorem that I've forgotten about. Commented Mar 12, 2016 at 0:00
• @alephzero . It is a standard theorem,and not hard.It has a companion : If $a_n\geq0$ then $\prod_n(1+a_n)<\infty \iff \sum_n a_n <\infty.$ Commented Mar 12, 2016 at 0:15

Your product is mostly understood here as : $$P^*= \lim_{n \to \infty} \prod_{k=1}^n (2k-1)/2k = \lim_{n \to \infty} \prod_{k=1}^n (1- \frac1{2k})$$

My intuition how to interpret your product was rather $$P= \lim_{n \to \infty} \prod_{k=1}^n k^{-(-1)^k}$$ and for this interpretation a regularization on the formal logarithm can be applied, namely Cearosummation $\mathfrak C$ for the logarithms of the product-terms and this gives indeed a relation to $\pi$: $$\log(P) = L \underset{\mathfrak C}=\sum_{k=1}^\infty (-1)^{k-1} \cdot \log(k) \approx -0.225791352645$$ The exponential is then $$P = \exp(L) = \sqrt{\frac2\pi}$$

Using Pari/GP:

L = -sumalt(k=1 , (-1)^k * log(k) )   \\ "sumalt()" provides sort of Cesarosum
print([ P=exp(L) , C = 1/sqrt(Pi/2)  , err = P - C ])
\\ result:
[0.797884560803, 0.797884560803, -1.005331151 E-200]


Remark: if that summation-method is allowed then the result is also expressible as the (regularized) derivative of the Dirichlet's eta-function at zero: (I've implemented the eta-function in Pari/GP using the Euler's conversion from the zeta:

$$L = - \eta(0)' \\ P = e^{- \eta(0)' }$$

While you evaluate your product as partial products $P_{1,n} = \frac12 \cdot \frac34 \cdots \frac {2n-1}{2n}$ one can do as well with $P_{2,n} = 1 \cdot \frac 32 \cdot \frac54 \cdots \frac{2n+1}{2n}$. While $\lim_{n \to \infty }P_{1,n}$ converges to zero diverges $P_{2,n}$ (to infinity).
But interestingly, if we take the partial evaluation up to the same $n$ , then we do $P = \lim_{n \to \infty} \sqrt{P_{1,n} \cdot P_{2,n}}$ with $P$ having the same value as given in my example using Cesaro-/Euler-summation.

We encounter here the same problem as when we sum the Grandi-series, with parentheses around pairs of terms in the two possible ways and get two different values: but when we take the mean of that two values we get the Cesaro-sum.

• note that $\eta'(0) = \lim_{s \to 0^+}\eta'(s)=\lim_{s \to 0^+} \sum_{n=1}^\infty n^{-s} (-1)^{n} \ln n$ so your regularization is $P = \lim_{s \to 0^+} \prod_{k=1}^\infty \frac{2k-1}{2k} \frac{e^{(2k-1)^{-s}}}{e^{(2k)^{-s}}}$ Commented Feb 13, 2017 at 16:34

Let $$a_n=\frac{(2n)!}{4^n(n!)^2}$$ then $$\frac1{a_n}=4^n(2n+1)B(n+1,n+1).$$ Hence $$\frac{1}{a_n}=(2n+1)2^{2n}\int_0^1 x^n(1-x)^ndx=\frac{2n+1}2\int_0^2u^n(2-u)^n du\\ >\frac{2n+1}{2}\int_{1-\frac{1}{\sqrt n}}^{1+\frac{1}{\sqrt n}}u^n(2-n)^ndu\\ >\frac{2n+1}{\sqrt n}(1-\frac1n)^n>\frac{2n+1}{4\sqrt n}$$ for $$n>1$$ and hence $$a_n<\frac{4\sqrt n}{2n+1}\rightarrow 0$$ as $$n\rightarrow\infty$$.

$$P_n=\prod_{i=1}^{n}\frac{(2i-1)}{2i}=\frac {\prod_{i=1}^{n}(2i-1) }{\prod_{i=1}^{n}(2i) }=\frac{\frac{2^n \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }} } {2^n \Gamma (n+1) }=\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }\,\, \Gamma (n+1)}$$

Take logarithms, use twice Stirling approximation and finish with Taylor series to get $$\log(P_n)=-\frac{1}{2} (\log (n)+\log (\pi ))-\frac{1}{8 n}+\frac{1}{192 n^3}-\frac{1}{640 n^5}+O\left(\frac{1}{n^7}\right)$$

Reuse Taylor since $$P_n=e^{\log(P_n)}$$ $$P_n=\frac{1}{\sqrt{\pi n}}\left(1-\frac{1}{8 n}+\frac{1}{128 n^2}+\frac{5}{1024 n^3}-\frac{21}{32768 n^4}-\frac{399}{262144 n^5}+O\left(\frac{1}{n^6}\right) \right)$$

Numerators and denominators form sequences $$A143503$$ and $$A061549$$ in $$OEIS$$.

For example, the above gives $$P_{10}=\frac{25888893921}{26214400000 \sqrt{10 \pi }}=\color{red}{0.17619705}19$$