Narrow Escape 1

Let $A_{k} = \frac{(k^2-1.21)}{(k^2-.81)}$ We are looking for $P = \prod_{k=2}^{\infty} A_{k}$ We will try and convert this into a telescoping product

$A_{k} = \frac{(k+1.1)(k-1.1)}{(k+0.9)(k-0.9)}$ Now let

$f(k) = k - 0.9$ $g(k) = k + 0.9$ ........ the product transforms as $P = \prod_{k=2}^{\infty} \frac{f(k+2)}{f(k)} \times \frac{g(k-2)}{g(k)}$ $P = ( \frac{f(4) . f(5) . f(6) .....}{f(2) . f(3) . f(4) ....} ) \times ( \frac{g(0) . g(1) . g(2) .....}{g(2) . g(3) . g(4) .....} )$

All that remains is $P = \frac{g(0) . g(1)}{f(2) . f(3)}$ $P = \frac{0. 9 \times 1.9}{1.1 \times 2.1}$ $P = \frac{57}{77}$

answer is $\boxed{134}$