Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$

Question

I needed to prove that

$\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$, $\forall n \geq 1$ .

I've atempted by induction.

I proved the case for $n=1$ and assumed it holds for some $n$.
The left-side of the n+1 case is
$\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2}$.

Using the inductive hypothesis, i could reach the result that

$\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} <\frac{2n+1}{\sqrt{2n+1}(2n+2)}=\frac{\sqrt{2n+1}}{(2n+2)}$.

Now, i'm wondering how should i connect it to my goal :

$\frac{1}{\sqrt{2n+2+1}}=\frac{1}{\sqrt{2n+3}} $

I know one way to prove that

$\frac{\sqrt{2n+1}}{(2n+2)}<\frac{1}{\sqrt{2n+3}} $

We just square things, then eventually reach

$ (n+1).(n+3)<(n+2)^2$

Which is easily provable because $3<4$ ....

But I was wondering if there was another way to show that ... perharps a more direct way to show that last bit ... a way that was not so direct and brute as to involve squaring both sides. A way of gradually manipulating the left-side until reaching the inequality with the right side.

Thanks in advance.

Answer 1

$$\begin{align}\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} &<\frac{2n+1}{\sqrt{2n+1}(2n+2)}\\~\\&=\frac{\sqrt{2n+1}}{(2n+2)}\\~\\&=\frac{\sqrt{(2n+1)(2n+3)}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{\sqrt{4n^2+8n+3}}{(2n+2)\sqrt{2n+3}}\\~\\&\lt \frac{\sqrt{4n^2+8n+3+\color{blue}{1}}}{(2n+2)\sqrt{2n+3}}\\~\\&= \frac{\sqrt{(2n+2)^2}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{1}{\sqrt{2n+3}}\end{align}$$

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Alternatively you may also use AM-GM inequality and conclude directly that $\sqrt{(2n+1)(2n+3)}\lt 2n+2$

Answer 2

Let $$a_n = \prod_{k=1}^n\frac{2k-1}{2k},\quad b_n=\prod_{k=1}^n\frac{2k}{2k+1}$$ then $0<a_n<b_n$ and so $a_n^2<a_nb_n=\frac{1}{2n+1}$.