# 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!

## 2 Answers

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.

• Thank you. So my attempt at 2. is wrong? – squenshl Aug 14 at 6:26
• It's not "wrong" per se... the only issue is that this sounds like a problem from the first or second chapter of an intro analysis course, would would be before we even talk about continuity, hence before differentiability. If that were correct, then it would be inappropriate to use a more advanced tool like L'Hopital's rule to solve it. – Brian Moehring Aug 14 at 6:33

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 $$n\geq 1$$.

• Great thank you! – squenshl Aug 14 at 6:25