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?
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 $1/ \sqrt \pi$
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, $$\lim A_n=0.$$
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.
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 $\quad \prod_{n=1}^{\infty}\frac {2 n-1}{2 n}=\prod_{n=1}^{\infty}(1-a_n)=0.$
Your product is mostly understood here as : $$P^*= \lim_{n \to \infty} \prod_{k=1}^n (2k-1)/2k = \lim_{n \to \infty} \prod_{k=1}^n (1- \frac1{2k}) $$
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:
$$ L = - \eta(0)' \\ P = e^{- \eta(0)' }$$
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.
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 $$P_{10}=\frac{25888893921}{26214400000 \sqrt{10 \pi }}=\color{red}{0.17619705}19$$
