<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 the induction hypotesis reduces to show that
$$ \frac{1}{\sqrt{2k+1}} \frac{(2k+1)}{2k+2}  ≤  \frac{1}{\sqrt{2k+3}} \implies\sqrt{(2k+1)(2k+3)}\le 2k+2$$
Can you finish from here?
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