Induction on sequences

I-Think · #1 $Old$ November 1st, 2009, 09:24 PM

The sequence of positive numbers $u_1, u_2, u_3,...$ is such that $u_1<4$
and
$u_{n+1}=\frac{5u_n+4}{u_n+2}$

By considering $4-u_{n+1}$ , prove by induction that $u_n<4$ for all $n\geq{1}$

My attempt N.B To make this easier to read I'm skipping all the official statements of the induction process
$4-u_{n+1}=\frac{4-u_n}{2+u_n}$

Let $n=1$
$4-u_2=\frac{4-u_1}{2+u_1}$
It is given that $u_1<4$ and positive so $\frac{4-u_1}{2+u_1}<4$
$u_2=4-\frac{4-u_1}{2+u_1}$
Hence $u_2$ is less than $4$

Assume true for $n=k$
$4-u_{k+1}=\frac{4-u_k}{2+u_k}$

Test at $n=k+1$

End of attempt

To finish this step confuses me. I implore the forum for assistance.

Jameson · #2 $Old$ November 2nd, 2009, 10:06 AM

Quote:

Originally Posted by I-Think $View Post$

The sequence of positive numbers $u_1, u_2, u_3,...$ is such that $u_1<4$
and
$u_{n+1}=\frac{5u_n+4}{u_n+2}$

By considering $4-u_{n+1}$ , prove by induction that $u_n<4$ for all $n\geq{1}$

My attempt N.B To make this easier to read I'm skipping all the official statements of the induction process
$4-u_{n+1}=\frac{4-u_n}{2+u_n}$

Let $n=1$
$4-u_2=\frac{4-u_1}{2+u_1}$
It is given that $u_1<4$ and positive so $\frac{4-u_1}{2+u_1}<4$
$u_2=4-\frac{4-u_1}{2+u_1}$
Hence $u_2$ is less than $4$

Assume true for $n=k$
$4-u_{k+1}=\frac{4-u_k}{2+u_k}$

Test at $n=k+1$

End of attempt

To finish this step confuses me. I implore the forum for assistance.

You're so close!

You have a nice way of writing 4-u_{k+1} in terms of u_k. So now you need to show that $\frac{4-u_k}{2+u_k}<4$ , which in turn shows that 4-u_{k+1}<4. This should be easy for you, since you have assumed that u_k < 4. Use that to show that the above fraction is < 4, thus 4-u_{k+1}<4, thus u_{k+1}<4. This shows that u_{k} < 4 implies u_{k+1}<4, which is the proof.