<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] Notice that 
\begin{align*}
a_n-a_{n+1}&=\frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}\\
&=\frac{1}{9^n}\left(\frac{9n^2-n^2-1-2n}{9}\right)\\
&=\frac{8n^2-1-2n}{9^{n+1}}.
\end{align*}
Now, $8n^2-1-2n=(4n+1)(2n-1)\gt 0$ because  $n\geq 1$.
</s>