# induction on a reccurence sequence

• November 29th 2008, 08:28 AM
induction on a reccurence sequence
i have a1=1

a(n+1) = (5an+4)/(an+2) for n>=1

need to prove by induction on n that 0<an<4 .

course of action would be?

then n is subscript and (n+1) also subscript , i think i need to get it into the form an+2 =A an+1 + B an

where A and B are some fixed numbers?

thanks
• November 29th 2008, 09:03 AM
running-gag
Hello !

The start of the recurrence is OK since 0 < a1=1 < 4

Suppose that $0 < a_n < 4$
Then
$a_{n+1}\;=\;\frac{5a_n+4}{a_n+2}$

$a_{n+1}\;=\;5-\frac{6}{a_n+2}$

$0 < a_n < 4$

$2 < a_n +2 < 6$

$\frac{1}{6} < \frac{1}{a_n+2} < \frac{1}{2}$

$1 < \frac{6}{a_n+2} < 3$

$2 < 5-\frac{6}{a_n+2} < 4$

$0 < 2 < a_{n+1} < 4$
• November 29th 2008, 09:24 AM