I've been stuck on this question for quite a while now.
$\forall m, n\in \mathbb{N}\, \text{we define}\, a_m = \sum^m_{k = 0} (\sqrt{k} + \sqrt{k+1})^{-1}\, , b_m = \sum^m_{k = 0} (\sqrt{k} + \sqrt{k+2})^{-1}\, and\, c_n= \prod^n_{m=0} \frac {a_m}{a_m-2(b_m-b_{m+1})}. \text{Find the smallest value of n} \in \mathbb{N}, \text{if any}, \text{for which}\, c_n < 0.01.$
I tried to rewrite $c_n$ algebraically but I was only able to come up with $$c_n= \prod^n_{m=0} \frac {a_m(\sqrt{m+1}+\sqrt{m+3})}{a_m(\sqrt{m+1}+\sqrt{m+3})-2}.$$ by simplifying $(b_m - b_{m+1})$ to $\frac{1}{\sqrt{m+1}+\sqrt{m+3}}$. I can't seem to do much with $a_m$ so I'm just not sure where to go from here or if I'm even in the right direction.
Forgive me if this is a bad post, I'm new to the website.
$a_m=\sum _{k=0}^m \dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\sum _{k=0}^m (\sqrt{k+1}-\sqrt{k})=\sqrt{m+1}$
$b_m=\sum _{k=0}^m \dfrac{1}{\sqrt{k+2}+\sqrt{k}}=\dfrac{1}{2}\,\sum _{k=0}^m (\sqrt{k+2}-\sqrt{k})=\dfrac{1}{2} \left(\sqrt{m+1}+\sqrt{m+2}-1\right)$
$b_m-b_{m+1}=\dfrac{1}{2} \left(\sqrt{m+1}-\sqrt{m+3}\right)$
$\dfrac {a_m}{a_m-2(b_m-b_{m+1})}=\dfrac{\sqrt{m+1}}{\sqrt{m+1}-\left(\sqrt{m+1}-\sqrt{m+3}\right)}=\dfrac{\sqrt{m+1}}{\sqrt{m+3}}$
$p(n)=\prod _{m=0}^n \dfrac{\sqrt{m+1}}{\sqrt{m+3}}=\dfrac{\sqrt{2} \sqrt{(n+1)!}}{\sqrt{(n+3)!}}=\color{blue}{\sqrt{\dfrac{2}{(n+2)(n+3)}}}$
Thanks to Steven Stadnicki for the blue simplification!
$ \begin{array}{l|l} n & p(n)\\ \hline 135 & 0.0102853 \\ 136 & 0.010211 \\ 137 & 0.0101378 \\ 138 & 0.0100656 \\ \color{red}{139} & \color{red}{0.0099945} \\ 140 & 0.00992437 \\ \end{array} $
Hope this can help