I am studying CIE A levels Further Maths and I am stuck at a question from June 2002:
Q
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 $4-u_{n+1}$, or otherwise, prove by induction that $u_n<4$ for all $n\geq1$.
Prove also that $u_{n+1}>u_n$ for all $n\geq 1$.
Here is my solution
To Prove the Initial Relation $u_n<4$
By using long division:
$$\frac{5u_n+4}{u_n+2}\equiv 5-\frac{6}{u_n+2}$$
For $n=1$
$$u_{2}= 5-\frac{6}{u_1+2}$$
Since $u_1<4$ then,
$$\frac{6}{u_1+2}>1$$
and,
$$5-\frac{6}{u_1+2}<4$$
therefore,
$$u_{2}<4$$
thus the statement is true for $n=1$
Conjecture: The statement is true for $n=k$
$$u_{k}<4$$
Since $u_{k+1}<4$ … (proved above for $n=1$),
then,
$$\frac{6}{u_{k+1}+2}>1$$
and,
$$5-\frac{6}{u_{k+1}+2}<4$$
therefore,
$$u_{k+2}<4$$
Since the statement is true for $n=k$, $n=k+1$, and $n=1$, the given relation ($u_n<4$) is correct.
To prove that $u_{n+1}>u_n$:
$$u_{n+1}-u_n=\frac{5u_n+4}{u_n+2}-u_n\equiv -\frac{(u_n)^2+3u_n+4}{u_2+2}$$
$$-\frac{(u_n)^2+3u_n+4}{u_2+2}\equiv \frac{(4-u_n)(u_n+1)}{u_n+2}>0$$
Since,
$$u_{n+1}-u_n>0$$
$$u_{n+1}>u_n$$
I am not sure if my solution to the first part is correct, and even if it is correct, I didn't use the relation $4-u_{n+1}$ so it would be really great if someone suggests abetter solution using the relation mentioned in the question.
As for second part, I am pretty confident that it is correct, but I am open to suggestions.
Best Answer
What they mean is that you should observe that $$4-u_{n+1}=4-\frac{5u_n+4}{u_n+2}=\frac{4-u_n}{u_n+2}$$ From this it is clear that if $4-u_n>0$, then $4-u_{n+1}>0$ and the gap gets smaller, so $u_n$ is increasing (to $4$ very fast).