How to show that bound?

Question

How to show that $\prod\limits_{i=1}^k \dfrac{k+i}{4i}$ is less than or equal to $1/2$ for all $k \ge 1$ Integer. I coulden't understand the answer in How to prove the bound on the probability?

Answer 1

Note $u_k=\prod\limits_{i=1}^k \dfrac{k+i}{4i}$. You have $\frac{u_{k+1}}{u_k}= \frac{2k+1}{2k+2}<1$. So $u_k$ is decreasing. The result then follows from $u_1=\frac{1}{2}$.

Answer 2

Use a simple mathematical induction.

${\displaystyle \prod_{i=1}^{k} \frac{k+i}{4i}}$ , $k\geq 1$

Basis: $k=1$

$${\displaystyle \prod_{i=1}^{1} \frac{1+i}{4i}}=\frac{2}{4} \leq \frac{1}{2}$$

Now show that if it holds for any fixed k it also holds for k+1.
Therefore assume $${\displaystyle \prod_{i=1}^{k} \frac{k+i}{4i}} \leq \frac{1}{2}$$

Inductive step: $k \rightarrow k+1$

$$ \begin{gather*} {\displaystyle \prod_{i=1}^{k+1} \frac{k+1+i}{4i}} \\ = {\displaystyle \prod_{i=1}^{k} \frac{k+i}{4i}} \times \frac{k+1+k}{4k} \leq \frac{1}{2} \times \frac{k+1+k}{4k} \\ = \frac{1}{2} \times \frac{2+1}{4} \\ = \frac{1}{2} \times \frac{3}{4} \leq \frac{1}{2} \end{gather*} $$

q.e.d.