Suppose a0>1 and an(read: a sub n)=2-1/a(sub n-1), n is an element of the natural numbers. Show that the sequence {a(sub n)} from n=1 to infinity is bounded and monotone. Find the limit.
Help, please!
$\displaystyle a_0>1. \ a_n = 2-\frac{1}{a_{n-1}}, n \in \{1, 2, ... \}$
It's almost always enlightening to simply write out the first few when given a recurrence formula:
$\displaystyle a_1 = 2-\frac{1}{a_{0}} = \frac{2a_0 - 1}{a_{0}}$
$\displaystyle a_2 = 2-\frac{1}{a_{1}} = 2 + \frac{-1}{\left(\frac{2a_0 - 1}{a_{0}}\right)} = 2 + \frac{-a_0}{2a_0 - 1} = \frac{3a_0 - 2}{2a_0 - 1}$
$\displaystyle a_3 = 2-\frac{1}{a_{2}} = 2 + \frac{-1}{\left(\frac{3a_0 - 2}{2a_0 - 1}\right)} = 2 + \frac{-2a_0 + 1}{3a_0 - 2} = \frac{4a_0 - 3}{3a_0 - 2}$
It's pretty obvious that the general form will be: $\displaystyle a_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)}, n \ge 1.$
Prove it by induction.
Then the monotinicity is just a matter of algebra, and the limit is easy to compute.
$\displaystyle \text{If }a_n = 2-\frac{1}{a_{n-1}}, n \in \{1, 2, ... \}, \text{ and } a_0>1$
$\displaystyle \text{then }a_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)} \text{ for all } n \ge 1.$
$\displaystyle \text{Proof: (by induction)}$
$\displaystyle \text{Define }b_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)} \text{ for all } n \ge 1.$
$\displaystyle \text{The claim then is that } a_n = b_n \ \forall n \ge 1.$
$\displaystyle \text{Have already shown that: }a_1 = \frac{2a_0 - 1}{a_{0}} = \frac{(1+1)a_0 - (1)}{(1)a_0 - (1-1)} = b_1,$
$\displaystyle a_2 = \frac{3a_0 - 2}{2a_0 - 1} = \frac{(2+1)a_0 - (2)}{(2)a_0 - (2-1)} = b_2,$
$\displaystyle \text{and } a_3 = \frac{4a_0 - 3}{3a_0 - 2} = \frac{(3+1)a_0 - (3)}{(3)a_0 - (3-1)} = b_3,$
$\displaystyle \text{so have established the claim for }n = 1, n=2, \text{ and } n=3 \text{ (that's more than is needed.)}$
$\displaystyle \text{Assume it's true for } n=k, \text{ for some } k \ge 1. \text{ Then }a_k = b_k.$
$\displaystyle \text{Thus }a_k = \frac{(k+1)a_0 - k}{ka_0 - (k-1)}.$
$\displaystyle \text{But by definition, have }a_{k+1} = 2-\frac{1}{a_{(k+1)-1}} = 2+\frac{-1}{a_k}.$
$\displaystyle \text{Thus }a_{k+1} = 2+\frac{-1}{\left(\frac{(k+1)a_0 - k}{ka_0 - (k-1)}\right)}$
$\displaystyle = 2+\frac{(-1)(ka_0 - (k-1))}{(k+1)a_0 - k}$
$\displaystyle = \frac{2((k+1)a_0 - k)}{(k+1)a_0 - k} + \frac{-ka_0 + (k-1)}{(k+1)a_0 - k}$
$\displaystyle = \frac{(2k+2)a_0 - 2k}{(k+1)a_0 - k} + \frac{-ka_0 + (k-1)}{(k+1)a_0 - k}$
$\displaystyle = \frac{[ (2k+2)a_0 - 2k ] + [ -ka_0 + (k-1) ] }{(k+1)a_0 - k}$
$\displaystyle = \frac{ (2k + 2 - k )a_0 + (- 2k + (k-1) ) }{(k+1)a_0 - k}$
$\displaystyle = \frac{ (k + 2)a_0 + (- k - 1) }{(k+1)a_0 - k}$
$\displaystyle = \frac{ ((k+1) + 1)a_0 - (k+1) }{(k+1)a_0 - ((k+1)-1)}$
$\displaystyle = b_{k+1}.$
$\displaystyle \text{Have proven that, for all }k \ge 1, a_k = b_k \text{ implies that } a_{k+1} = b_{k+1}.$
$\displaystyle \text{Since }a_1 = b_1, \text{ that proves by induction that } a_n = b_n \text{ for all } n \ge 1,$
$\displaystyle \text{and so have proven that } a_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)} \text{ for all } n \ge 1.$