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!!!
 A: 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 $0$.
A: Note that for  $ n\ge 3$, we have $$2^{n^2} >6^n$$
Therefore  $$\frac{5^n}{2^{n^2}}<(5/6)^n \to 0$$
A: $\dfrac {a_{n+1}}{a_n}=\frac 52\times 4^{-n}\to 0$ so if $a_n\to \ell$ and $|\ell|>0$ 
Then $\frac{\ell}{\ell}=0$ by limit identification, this is a contradiction, so $\ell=0$
Alternately for $n\ge 1,\quad\frac 52\times 4^{-n}\le \frac 58<1$ 
Thus $0\le a_n\le\left(\frac 58\right)^{n-1}a_1\to 0$
Note: You should have seen this from question 2. $a_{n+1}-a_n<0$ is not easy to study, but $\frac{a_{n+1}}{a_n}<1$ is, both giving you the decreasing behaviour. Consequently you get a geometric sequence with a reason lower than $1$, which is forced to converge to $0$. You missed this point.
A: 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 $\lim_{n\to\infty} 2^{2n+1} = \infty$.
