<s>[INST] I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:

$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k}  ≤  \frac{1}{\sqrt{2k+1}}$

Let's prove that the unequality is right for $n = k + 1$ as well, that is:

$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2}  ≤  \frac{1}{\sqrt{2k+3}}$

I've spent a comparable amount of time trying to prove this point but looks like I can't grasp what to do.
 [/INST] Note that if $$\frac{a}{b}<1 \Rightarrow \frac{a}{b} < \frac{a+x}{b+x}, x>0.~~~(1)$$
Let  $$P_1=\frac{1}{2}~ \frac{3}{4}~ \frac{5}{6}....\frac{2k+1}{2k+2}.$$
Let us add 1, up and down in each fraction to get
$$P_2=\frac{2}{3} ~\frac{4}{5} ~\frac{6}{7}...\frac{2k+2}{2k+3}.$$
So By (1), $$P_1<P_2 \Rightarrow P_1^2 < P_1. P_2 \Rightarrow P_1^2 < \frac{1}{2k+3}$$$$\Rightarrow P_1 <\frac{1}{\sqrt{2k+3}}.$$
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