Finding $ \lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$

Question

Finding

$$\lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$$

Suspecting that the limit is 0, but how do I show this? Was able to get an upper bound of $ \frac{1}{2\sqrt{e}} $ easily but it's not useful. Apparently this is a hard problem.

Answer 1

You may write $$ \begin{align} \frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n} &=\frac{1\cdot \color{blue}2\cdot 3\cdot \color{blue}4 \cdot 5\cdot\color{blue}6\cdots(2n-1)\cdot \color{blue}{2n}}{(2\cdot 4\cdot 6\cdots (2n))^\color{blue}2}\\ &=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2}\\ & =\frac{(2n)!}{2^{2n} (n!)^2 }\\ & =\frac{1}{\sqrt{\pi n}}+\mathcal{O}\left(\frac{1}{n^{3/2}}\right), \quad \text{for} \, n \, \text{great} \end{align} $$ where we have used Stirling's approximation (approximating $(2n)!$ and $n!$), then you easily conclude for the limit (here you have more than the limit).

Answer 2

Compare $A=\prod_{j=1}^n\frac{2j-1}{2j}$ with $B=\prod_{j=1}^n\frac{2j}{2j+1}$.
Each factor in A is less than the corresponding factor in B so $A<B$.
AB = $1/(2n+1)$ by the telescope principle.
$A^2<AB=1/(2n+1)$
$\lim_{n\to\infty}A=0$

Answer 3

I think that Stirling's approximation is an overkill. Notice that for any $j>1$: $$\left(1-\frac{1}{2j}\right)^2 = 1-\frac{1}{j}+\frac{1}{4j^2} = \left(1-\frac{1}{j}\right)\cdot \left(1+\frac{1}{4j(j-1)}\right)\tag{1}, $$ so given that $P_N = \prod_{j=1}^{N}\left(1-\frac{1}{2j}\right)$, we have: $$ P_N^2 = \frac{1}{4}\prod_{j=2}^{N}\left(1-\frac{1}{j}\right)\prod_{j=1}^{N-1}\left(1+\frac{1}{4j(j+1)}\right)=\frac{1}{4N}\prod_{j=1}^{N-1}\left(1+\frac{1}{4j(j+1)}\right).\tag{2}$$ Since $1+x<e^x$ for any $x\neq 0$, the last product can be bounded by: $$\prod_{j=1}^{N-1}\left(1+\frac{1}{4j(j+1)}\right)\leq \exp\left(\frac{1}{4}\sum_{n=1}^{N-1}\frac{1}{n(n+1)}\right)\leq e^{\frac{1}{4}},\tag{3}$$ so we get the inequality: $$ P_N \leq \frac{1}{2\sqrt{N}}e^{\frac{1}{8}} \tag{4} $$ that is enough to state $\lim_{N\to +\infty} P_N = 0$, as wanted.

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As an alternative approach, it is worth to recall that the infinite product $$\prod(1+a_n)$$ is convergent iff $\sum a_n$ is convergent. Since $$\sum_{n\geq 1}\frac{1}{2n-1}$$ is divergent, the product $$\prod_{n\geq 1}\left(\frac{2n}{2n-1}\right)$$ is divergent, too, so your product converges towards zero.