calculus - Proving the convergence of the series $sum_{j=1}^infty frac{(2^j)^2}{j!}$ without root or ratio test

I am given the series:
$\displaystyle \sum_{j=1}^\infty \frac{(2^j)^2}{j!}$ and I can show that it converges by using the ratio test, but I'm not sure how to approach to prove its convergence without it.

Answer

Let
$a_j =\frac{(2^j)^2}{j!}$.
Then
$\frac{a_{j+1}}{a_j} =\frac{\frac{(2^{j+1})^2}{(j+1)!}}{\frac{(2^j)^2}{j!}} =\frac{2^{2(j+1)-2j}}{j+1} =\frac{4}{j+1}$.

Therefore,
for
$j \ge 9$,
$\frac{a_{j+1}}{a_j} \le \frac12$.

By induction,
for

$j \ge 9$ and
$k \ge 1$,
$\frac{a_{j+k}}{a_j} \le \frac1{2^k}$.

Therefore
for
$j \ge 9$ and
$k \ge 1$,

$a_{j+k} \le a_j\frac1{2^k}$
so that,
for $j \le 9$,
$\sum_{k=1}^{\infty} a_{j+k} \le \sum_{k=1}^{\infty} a_j\frac1{2^k} =a_j$.

Since initial terms of a sum
do not affect the convergence,
the sum converges.