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!!!

For the first question: The are many ways to show this but we will use induction. Well $a_1 = \frac{5}{2} > 0$. Now suppose that $a_n > 0$ for some $n$. Then
\begin{equation*}
a_{n+1} = \frac{5^{n+1}}{2^{(n+1)^2}} = \frac{5^{n+1}}{2^{n^2+2n+1}} = \frac{5}{2^{2n+1}}\cdot \frac{5^n}{2^{n^2}} = \frac{5}{2^{2n+1}}\cdot a_n.
\end{equation*}
Since $a_n > 0$ and $\frac{5}{2^{2n+1}} > 0$, we see that $a_{n+1} > 0$. Hence, $a_n > 0$ as required.
For the second question: We have
\begin{equation*}
a_{n+1}-a_n = \frac{5}{2^{2n+1}}\cdot a_n-a_n = \left(\frac{5}{2^{2n+1}}-1\right)a_n.
\end{equation*}
Now $a_n > 0$ so
\begin{equation*}
a_{n+1}-a_n < 0 \Longleftrightarrow \frac{5}{2^{2n+1}}-1 < 0 \Longleftrightarrow 5 < 2^{2n+1},
\end{equation*}
which is easy to prove by induction: For $n = 1$, $2^{2\times 1+1} = 2^3 = 8 > 5$. Now if $2^{2n+1} > 5$, then
\begin{equation*}
2^{2(n+1)+1} = 2^{2n+3} = 2^2\cdot 2^{2n+1} = 4\cdot 2^{2n+1} > 4\times 5 = 20 > 5.
\end{equation*}
Thus, the sequence $(a_n)$ is strictly decreasing as required.
For the final question: Since we have a (monotonically) decreasing sequence which is bounded below, by the monotone convergence theorem $(a_n)$ converges, say to $\ell$. Then because $a_{n+1} = \frac{5}{2^{2n+1}}\cdot a_n$ we get
\begin{equation*}
\begin{split}
\ell = \left(\lim_{n\to\infty} \frac{5}{2^{2n+1}}\right)\cdot \ell = 0\cdot \ell = 0
\end{split}
\end{equation*}
because $\lim_{n\to\infty} 2^{2n+1} = \infty$.
