Challenging Product

We factor the term

$\dfrac{n^2 + 13n}{n^2 + 13n + 40} = \dfrac{n (n + 13)}{(n + 5)(n + 8)}$

and then patiently write out the first several terms until we notice cancellations:

$\dfrac{1 \cdot \not{14}}{\not{6} \cdot 9} \cdot \dfrac{2 \cdot \not{15}}{\not{7} \cdot 10} \cdot \dfrac{3 \cdot \not{16}}{\not{8} \cdot 11} \cdot \dfrac{4 \cdot 17}{9 \cdot 12} \cdot \dfrac{5 \cdot 18}{10 \cdot 13} \cdot \dfrac{\not{6} \cdot 19}{11 \cdot \not{14}} \cdot \dfrac{\not{7} \cdot 20}{12 \cdot \not{15}} \cdot \dfrac{\not{8} \cdot 21}{13 \cdot \not{16}} \cdot \ ....$

At that point it's not difficult to see that the only factors to survive are $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5$ in the numerators and $9 \cdot 10 \cdot 11 \cdot 12 \cdot 13$ in the denominators, so the infinite product converges to

$\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}{9 \cdot 10 \cdot 11 \cdot 12 \cdot 13} = \dfrac{1}{1287}$

and our answer is $1 + 1287 = \boxed{1288}$