Inequation of an sum smaller than 1

Question

I'm trying to figure out the following $$ \sum^{\infty}_{n=3} \dfrac{q!^2}{n!^2} < 1 $$ How I can show it if $q \geq 2$?

Maybe with telescoping sums?

Thanks,

Landau

Answer 1

We have $$\sum_{n=0}^{\infty} \dfrac1{(n!)^2} = I_0(2)$$ where $I_0(z)$ is the modified Bessel function of the first kind. In your problem, $$\sum_{n=3}^{\infty} \dfrac1{(n!)^2} = I_0(2) - \dfrac94 \approx 0.029585302336067267$$ Hence, we get $$(q!)^2 < 34 \implies q! < 6 \implies q < 3$$

Answer 2

Provided that $q\leq 3$, each term of the sum is less than or equal to $1$ and is therefore less than or equal to its square root. Using the power series expansion for $e$, we find that $$ \sum_{n=3}^\infty \frac{q!^2}{n!^2} < \sum_{n=3}^\infty \frac{q!}{n!} = q!\Big(e-\frac{5}{2}\Big), $$ and this expression is less than $1$ when $q\leq 2$