calculus - Limit of $a_n := frac{5^n}{2^{n^2}}$

Consider the sequence $(a_n)$ defined by $a_n := \frac{5^n}{2^{n^2}}$.
1. Prove that the sequence $(a_n)$ is bounded below by $0$.
We note that $a_n > 0$ for $n\geq 0$. Thus, the sequence is bounded from below.
2. Prove that the sequence $(a_n)$ is strictly decreasing by showing that $a_{n+1}-a_n < 0$ for all $n\in \mathbb{N}$.
We look to $a_n = \frac{5^n}{2^{n^2}}$ and $a_{n+1} = \frac{5^{n+1}}{2^{(n+1)^2}}$. For $n\geq 1$ we see that $a_n > a_{n+1}$. Therefore, we have a strictly decreasing sequence.
3. Deduce that the sequence $(a_n)$ converges and calculate its limit.
Since we have a (monotonically) decreasing sequence which is bounded below, by the monotone convergence theorem this sequence converges. How do we find the limit? Is it the squeeze theorem? Thank you for the help!!!

Answer

Once you know a limit $L$ exists, then find a recurrence relation, like
$$a_{n+1} = \frac52\frac{1}{2^{2n}}a_n$$
And take the limit as $n\to\infty$:
$$L = \frac{5}{2}\cdot0\cdot L$$
which implies that limit $L$ must equal $0$.