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.