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. 
 A: It can be done by explicit comparison with some nice series. For the sum from $8$ to $\infty$, note that the top is $4^j$ and the bottom is greater than $8^{j-7}$, that is, $8^{-7}8^j$. Thus for $j\ge 8$ we can compare with $8^7\sum_8^\infty \left(\frac{4}{8}\right)^j$. 
Remark: This comparison is in principle not far from the Ratio Test, which fundamentally is also a comparison with a geometric series.
A: 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.
A: We have: $\dfrac{j^2}{j!}= \dfrac{j}{(j-1)!}= \dfrac{1}{(j-2)!}+ \dfrac{1}{(j-1)!}$, and you can reindex to show each of them converges, and they are easy to show to be convergent by comparing each with a $p>1$ -series.
If you have the series with $2^{2j} = 4^j$, then you would use $e^4$ as the targeting sum for the series. So either series is convergent hand down.
A: If you know $\sum 1/j^2 < \infty,$ you can argue like this: For $j\ge 6,$
$$\frac{4^j}{j!} = \frac{4}{j}\frac{4}{j-1} \cdots \frac{4}{4}\frac{4}{3}\frac{4}{2}\frac{4}{1}\le  \frac {4^6}{4!}\frac{1}{j(j-1)}.$$
Dividing the term on the right by $1/j^2$ and letting $j\to \infty$ gives a finite limit. Thus the limit comparison test shows our series converges.
