Prove that $\displaystyle a_n=\frac{1\cdot 3\cdots(2n+1)}{2\cdot 4\cdots(2n)}$ is not bounded above.
We use $\displaystyle \ln(1+x)\geq x-\frac{x^2}2$ for $\displaystyle x\geq 0$ to get $\displaystyle \ln a_n=\sum_{j=1}^n\ln\left(1+\frac 1{2j}\right)\geq \sum_{j=1}^n\frac 1{2j}-\frac 12\sum_{j=1}^n\frac 1{(2j)^2}=\frac 12\sum_{j=1}^n\frac 1j-\frac 18\sum_{j=1}^n\frac 1{j^2}$.
Since $\displaystyle \lim_{n\to \infty}\sum_{j=1}^n\frac 1j=+\infty$ and $\displaystyle \lim_{n\to \infty}\sum_{j=1}^n\frac 1{j^2}<+\infty$, we get the result.