# The sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$

1. 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.
2. 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!

1. 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 $$\frac{a_{n+1}}{a_n} = \frac{1}{9}\left(1 + \frac{1}{n}\right)^2 \leq \frac{1}{9}\left(1 + \frac{1}{1}\right)^2 = \frac{4}{9} < 1$$
2. 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.
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 $$n\geq 1$$.