showing the limit of a recurrence relation The recurrence relation is defined as such:
$$a_n=2+\frac{80}{a_{n-1}}$$
It is also given that $a_1=2$, how do we show that $$\lim_{n\to\infty}a_n=10 ?$$
I am totally stuck at how I should approach this question. 
Thanks in advance for any help or suggestion!
 A: It can be shown easily by mathematical induction that (cf. Mr. Scott's answer) the subsequence $\{a_{2n}\}_{n \geq 1}$ is decreasing and bounded below by $10$, whereas the subsequence $\{a_{2n - 1}\}_{n \geq 1}$ is increasing and bounded above by $10$, hence by monotone bounded convergence theorem, each subsequence converges, denote their limits by $\alpha$ and $\beta$ respectively, which are all strictly greater than $0$. On one hand, let $n \to \infty$ in 
$$a_{2n + 1} = 2 + \frac{80}{a_{2n}}$$
to get 
$$\beta = 2 + \frac{80}{\alpha}.\tag{1}$$
On the other hand, let $n \to \infty$ in 
$$a_{2n + 2} = 2 + \frac{80}{a_{2n + 1}}$$
to get 
$$\alpha = 2 + \frac{80}{\beta}. \tag{2}$$
Solve $(1)$ and $(2)$ for $\alpha, \beta$, we have (the negative roots are discarded)
$$\alpha = \beta = 10.$$
showing that $\{a_n\}$ converges (using the well-known proposition that a sequence converges if and only if its odd-subsequence and even-subsequence converge to the same value), and the limit is $10$.
P.S. In response to David C. Ullrich's concern (it's a very good one, we should not leave out any nontrivial details), I will add the details showing $\{a_{2n}\}$ and $\{a_{2n - 1}\}$ are monotonic. The bounded part has already been shown in Mr.Scott's answer.
For $n = 1$, $a_1 = 2, a_2 = 2 + 80/2 = 42, a_3 = 2 + 80/43 > 2, a_4 < 2 + 80/2 = 42 = a_2$. The initial steps have been checked.
Assume for $n = k > 1$, we have $a_{2k} < a_{2k - 2}, a_{2k + 1} > a_{2k - 1}$. Then for $n = k + 1$, 
$$a_{2k + 2} - a_{2k} = (2 + 80/a_{2k + 1}) - (2 + 80/a_{2k - 1}) = 80(1/a_{2k + 1} - 1/a_{2k - 1}) < 0$$
by induction assumption, and
$$a_{2k + 3} - a_{2k + 1} = (2 + 80/a_{2k + 2}) - (2 + 80/a_{2k}) = 80(1/a_{2k + 2} - 1/a_{2k}) > 0$$ 
by above inequality just proved. Hence the induction step is completed. 
A: Use the recurrence
$$a_n=2+\frac{80}{a_{n-1}}$$
to show that if $a_{n-1}<10$, then $a_n>10$, while if $a_{n-1}>10$, then $a_n<10$. Since $a_1=2<10$, we have $a_n<10$ when $n$ is odd, and $a_n>10$ when $n$ is even. Now
$$\begin{align*}
|a_n-10|&=\left|\frac{80}{a_{n-1}}-8\right|\\
&=\left|\frac{80-8a_{n-1}}{a_{n-1}}\right|\\
&=\left|\frac{8(10-a_{n-1})}{a_{n-1}}\right|\\
&=\left|\frac8{a_{n-1}}\right|\cdot|a_{n-1}-10|\;,
\end{align*}$$
so $|a_n-10|<|a_{n-1}-10|$ provided that $a_{n-1}>8$. Suppose that you can find an $m\in\Bbb Z^+$ such that $8<a_m<10$. 


*

*Show that $a_n>8$ for all $n\ge m$.

*Let $r=\frac8{a_m}$, and show by induction on $k$ that $$|a_{m+k}-10|\le r^k|a_m-10|$$ for all $k\ge 0$. Do you see why that implies the desired result?
All that’s needed now is to find such an $m$; it’s a little tedious, but you can do it by direct calculation; the smallest such $m$ is less than $15$.
A: What Zhanxiong wrote looks simpler, but it looks like maybe it just looks simpler because he left out some details. Someone should point out that no, looking at it that way makes the details simpler as well.
The recurrence shows that $$a_{n+2}-10=\frac{32}{40+a_n}(a_n-10).$$
So all we have to show is that $a_n>0$ and we know that the numbers $a_{2n}-10$ all have the same sign and decrease geometrically in absolute value; similarly for $a_{2n+1}$.
