Bounded and Monotone

**lovesmath** · October 17th 2012, 03:09 PM

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!

**johnsomeone** · October 17th 2012, 03:23 PM

$a_0>1. \ a_n = 2-\frac{1}{a_{n-1}}, n \in \{1, 2, ... \}$

It's almost always enlightening to simply write out the first few when given a recurrence formula:

$a_1 = 2-\frac{1}{a_{0}} = \frac{2a_0 - 1}{a_{0}}$

$a_2 = 2-\frac{1}{a_{1}} = 2 + \frac{-1}{\left(\frac{2a_0 - 1}{a_{0}}\right)} = 2 + \frac{-a_0}{2a_0 - 1} = \frac{3a_0 - 2}{2a_0 - 1}$

$a_3 = 2-\frac{1}{a_{2}} = 2 + \frac{-1}{\left(\frac{3a_0 - 2}{2a_0 - 1}\right)} = 2 + \frac{-2a_0 + 1}{3a_0 - 2} = \frac{4a_0 - 3}{3a_0 - 2}$

It's pretty obvious that the general form will be: $a_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)}, n \ge 1.$

Prove it by induction.

Then the monotinicity is just a matter of algebra, and the limit is easy to compute.

**lovesmath** · October 21st 2012, 01:31 PM

I am having trouble proving by induction. How should I set that up?

**johnsomeone** · October 21st 2012, 03:07 PM

$\text{If }a_n = 2-\frac{1}{a_{n-1}}, n \in \{1, 2, ... \}, \text{ and } a_0>1$

$\text{then }a_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)} \text{ for all } n \ge 1.$

$\text{Proof: (by induction)}$

$\text{Define }b_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)} \text{ for all } n \ge 1.$

$\text{The claim then is that } a_n = b_n \ \forall n \ge 1.$

$\text{Have already shown that: }a_1 = \frac{2a_0 - 1}{a_{0}} = \frac{(1+1)a_0 - (1)}{(1)a_0 - (1-1)} = b_1,$

$a_2 = \frac{3a_0 - 2}{2a_0 - 1} = \frac{(2+1)a_0 - (2)}{(2)a_0 - (2-1)} = b_2,$

$\text{and } a_3 = \frac{4a_0 - 3}{3a_0 - 2} = \frac{(3+1)a_0 - (3)}{(3)a_0 - (3-1)} = b_3,$

$\text{so have established the claim for }n = 1, n=2, \text{ and } n=3 \text{ (that's more than is needed.)}$

$\text{Assume it's true for } n=k, \text{ for some } k \ge 1. \text{ Then }a_k = b_k.$

$\text{Thus }a_k = \frac{(k+1)a_0 - k}{ka_0 - (k-1)}.$

$\text{But by definition, have }a_{k+1} = 2-\frac{1}{a_{(k+1)-1}} = 2+\frac{-1}{a_k}.$

$\text{Thus }a_{k+1} = 2+\frac{-1}{\left(\frac{(k+1)a_0 - k}{ka_0 - (k-1)}\right)}$

$= 2+\frac{(-1)(ka_0 - (k-1))}{(k+1)a_0 - k}$

$= \frac{2((k+1)a_0 - k)}{(k+1)a_0 - k} + \frac{-ka_0 + (k-1)}{(k+1)a_0 - k}$

$= \frac{(2k+2)a_0 - 2k}{(k+1)a_0 - k} + \frac{-ka_0 + (k-1)}{(k+1)a_0 - k}$

$= \frac{[ (2k+2)a_0 - 2k ] + [ -ka_0 + (k-1) ] }{(k+1)a_0 - k}$

$= \frac{ (2k + 2 - k )a_0 + (- 2k + (k-1) ) }{(k+1)a_0 - k}$

$= \frac{ (k + 2)a_0 + (- k - 1) }{(k+1)a_0 - k}$

$= \frac{ ((k+1) + 1)a_0 - (k+1) }{(k+1)a_0 - ((k+1)-1)}$

$= b_{k+1}.$

$\text{Have proven that, for all }k \ge 1, a_k = b_k \text{ implies that } a_{k+1} = b_{k+1}.$

$\text{Since }a_1 = b_1, \text{ that proves by induction that } a_n = b_n \text{ for all } n \ge 1,$

$\text{and so have proven that } a_n = \frac{(n+1)a_0 - n}{na_0 - (n-1)} \text{ for all } n \ge 1.$

**lovesmath** · October 21st 2012, 03:55 PM

This helped so much. I was able to see where I went wrong. Thanks!

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