1. ## Help! Sequence question

For the sequence a1= 1, a(n+1)= 1 + 2a(n)/ 1 + a(n) n=1,2,3,..

we need to show that the sequence is bounded from above by 100.

I talked to my prof and he said to use induction with a(k) < 100 to show a(k+1) < 100.

if I set it up, i get
1 + 2a(k)/ 1+ a(k) < 100
1 + 2a(k) < 100 (1+ a(k))
1 + 2a(k) < 100 + 100 a(k)
-99 < 98 a(k)

which is where I am stuck. How would I use this to show that a(n) is bounded by 100?

2. Hello

Ok, I'll just sketch the proof for you.

n=1
$\displaystyle a_1 = 1 < 100$

Therefore true for n=1, assume n=k is true, that is,

$\displaystyle a_{k+1} = \frac{1+2a_k}{1+a_k} < 100$

Consider n=k+1

$\displaystyle a_{k+2} = \frac{1+2a_{k+1}}{1+a_{k+1}}$

$\displaystyle = \frac{1 + \frac{2+4a_k}{1+a_k}}{1 + \frac{1+2a_k}{1+a_k}}$

after a bit of simple algebra, you'll arrive at

$\displaystyle a_{k+2} = \frac{3+5a_k}{2+3a_k} < \frac{3+6a_k}{3+3a_k}$ since $\displaystyle a_k > 1$, $\displaystyle \forall k \in N$

But $\displaystyle \frac{3 + 6a_k}{3+3a_k} = \frac{1 + 2a_k}{1+ a_k} < 100$ by the induction hypothesis.

so $\displaystyle a_{k+2} < 100$

Therefore, if n=k is true, then n=k+1 is true. But since n=1 is true, this implies than n=2 is true, which implies n=3 is true..etc. Hence $\displaystyle a_{n+1} < 100$ for all natural numbers n, which, by definition, implies that 100 is an upper bound.

3. The only problem with that proof is that it is stating that ak+2<ak+1 when in fact ak+2 > ak+1 because {an} is increasing

4. $\displaystyle a_{k+2} = \frac{3+5a_k}{2+3a_k} < \frac{3+6a_k}{3+3a_k} since a_k > 1, \forall k \in N$
The inequality here doesn't work eg $\displaystyle a_k = 1$.

I'm not sure what approach your professor was expecting but here is a method of proof:

$\displaystyle a_{k+1} = \frac {1+2a_k}{1+a_k}$
$\displaystyle =\frac {2a_k+2-1}{a_k+1}$
$\displaystyle = 2 - \frac {1}{a_k+1}$
<2 since $\displaystyle a_k>-1.$
so $\displaystyle a_k<100$