I was requested to determine if $$a_n=\frac{2^n}{n!}$$ is convergent or not. Please notice that I was not requested to present a formal proof that the sequence is convergent, but only to determine if it was, so "intuitive" approaches ar valid as long as I can justify them.
What I did was the following.
$i)$ Notice that $$a_n=\frac{2^n}{n!}=\frac{2*2*2*...*2}{1*2*3*...*n}=\frac{2}{1}*\frac{2}{2}*\frac{2}{3}*...*\frac{2}{n}$$.
$ii$) From $i$ it follows that
$$lim_{n\to\infty}a_n=lim_{n\to\infty}(\frac{2}{1}*\frac{2}{2}*\frac{2}{3}*...*\frac{2}{n})$$
$$= lim_{n\to\infty}(\frac{2}{1})*lim_{n\to\infty}(\frac{2}{2})*lim_{n\to\infty}(\frac{2}{3})*...*lim_{n\to\infty}(\frac{2}{n})$$
Notice that we can be sure the limit of each of the terms of the sequence represented by $...$ exists when $n\to\infty$ because they are all constants of the form $\frac{2}{b}$ with $b\in\mathbb{N}$.
$iii$) Solving $ii$ we get $$2*1*2/3*...*0=0$$
Because $lim_{n\to\infty}a_n=0$ the sequence converges.
Yes, the limit is $0$, but you argument is not correct. To see why, consider the sequence$$1,\left(1+\frac12\right)\times\left(1+\frac12\right),\left(1+\frac13\right)\times\left(1+\frac13\right)\times\left(1+\frac13\right),\ldots$$In other words, consider the sequence$$\left(\left(1+\frac1n\right)^n\right)_{n\in\mathbb N}.$$Its limit is $e$, right?! But, from your argument, you should deduce that the limit is $1$, since $\lim_{n\to\infty}1+\frac1n=1$.
However, as I wrote, the limit is indeed $0$. If $n>2$, then $\frac 2n\leqslant\frac23$. Therefore$$n>2\implies\frac21\times\frac22\times\frac23\times\cdots\times\frac2n\leqslant2\times\left(\frac23\right)^{n-2}.$$