<s>[INST] The sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$ 
*

*Show that the sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$ is monotone decreasing and bounded below.
Let $a_n = \frac{n^2}{9^n}$. For $n\geq 1$ we have
\begin{equation*}
a_n-a_{n+1} = \frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}.
\end{equation*}
To show this sequence is decreasing we want to show that this quantity is positive. Now we know that $n^2 > 0$, and $(n+1)^2 > 0$ for $n\geq 1$ so $a_n-a_{n+1} > 0$, i.e. $a_n > a_{n+1}$. Since it is decreasing it must be monotone decreasing as required.
A sequence is bounded below if there is a number $m$ such that $m\leq a_n$ for all $n$. Since all the terms in this sequence are all positive, the sequence is bounded below by $0$, i.e. $a_n\geq 0$ as required.  

*Use the result in (i) to prove that $\lim_{n\to\infty} \frac{n^2}{9^n} = 0$.
Since this sequence is monotone decreasing and bounded below, it converges by the monotone convergence theorem. Lets re-write this a little. We can do this by writing
\begin{equation*}
\frac{n^2}{9^n} = \frac{n^2}{\left(e^{\ln{(9)}}\right)^n} = \frac{n^2}{e^{n\ln{(9)}}}
\end{equation*}
Now we have a limit of the form $\frac{\infty}{\infty}$ so we can apply L'Hopital's rule which gives
\begin{equation*}
\begin{split}
\lim_{n\to\infty} \frac{n^2}{9^n} &= \lim_{n\to\infty} \frac{n^2}{e^{n\ln{(9)}}} \\
&= \lim_{n\to\infty} \frac{2n}{\ln{(9)}e^{n\ln{(9)}}} \\
&= \lim_{n\to\infty} \frac{2}{\left(\ln{(9)}\right)^2e^{n\ln{(9)}}} \\
&= 0
\end{split}
\end{equation*}
as required.
Is this good? Thanks!

 [/INST] *

*First note that $a_n > 0$ for all $n.$  Therefore $$a_{n+1} \leq a_n \iff \frac{a_{n+1}}{a_n} \leq 1$$ and in our case, we have $$\frac{a_{n+1}}{a_n} = \frac{1}{9}\left(1 + \frac{1}{n}\right)^2 \leq \frac{1}{9}\left(1 + \frac{1}{1}\right)^2 = \frac{4}{9} < 1$$

*We know that the limit exists, so write $L = \displaystyle\lim_{n\to\infty} a_n$ and suppose $L > 0.$  Then there will be some $N$ such that $n \geq N$ implies $$a_n - L < L \\ \implies \\ a_n < 2L \\ \implies \\  a_{n+1} = a_n \cdot \left(\frac{a_{n+1}}{a_n}\right) < (2L) \cdot \left(\frac{4}{9}\right) < L$$
which is a contradiction since a monotone decreasing sequence cannot ever be less than its limit.
</s>