# Proving that a sequence is increasing

Problem: A sequence $(a_n)$ is defined recursively as follows, where $0<\alpha\leqslant 2$: $$a_1=\alpha,\quad a_{n+1}=\frac{6(1+a_n)}{7+a_n}.$$ Prove that this sequence is increasing and bounded above by $2$. What is its limit?

Ideas: How should I go about starting this proof? If the value of $a_1$ were given, I could show numerically and by induction that the sequence is increasing. But no exact value of $\alpha$ is given.

• Just simply assume you "know" the $\alpha$ value. – Daniel Mar 28 '15 at 5:44
• No, I'm saying that the value of $\alpha$ is irrelevant. – Daniel Mar 28 '15 at 5:46
• I don't think I've ever seen a chosen answer at $-3$ with another answer provided at $+6$. Fascinating. – Daniel W. Farlow Mar 28 '15 at 6:15
• @science What advice did I give? Also, downvotes do mean things in general. They can trigger system flags, they communicate a worthless or flawed answer, etc. They're not for nothing. – Daniel W. Farlow Mar 28 '15 at 7:32
• science's answer is completely incorrect, and Nathanson had already pointed out a counter-example in his comment. Besides learning to solve math questions, you also need to learn to ascertain the correctness of any answer in general, otherwise you will never know for sure what is correct. – user21820 Mar 29 '15 at 10:37

First, $0\leq a_1\leq2$.

Second, $a_{n+1}=\frac{6(1+a_n)}{7+a_n}=6-\frac{36}{7+a_n}$.

So, if $0\leq a_n\leq2$,

$\qquad$then $7\leq 7+a_n\leq 9$,

$\qquad$then $\frac{1}{7}\geq\frac{1}{7+a_n}\geq\frac{1}{9}$,

$\qquad$then $\frac{36}{7}\geq\frac{36}{7+a_n}\geq\frac{36}{9}$,

$\qquad$then $-\frac{36}{7}\leq-\frac{36}{7+a_n}\leq-\frac{36}{9}$,

$\qquad$then $0\leq6-\frac{36}{7}\leq6-\frac{36}{7+a_n}\leq6-\frac{36}{9}=2$,

then $0\leq a_{n+1}\leq2$.

By induction $0\leq a_n\leq2$ for all $n$.

Also $a_{n+1}-a_n=\frac{6(1+a_n)}{7+a_n}-a_n=\frac{6-a_n-a_n^2}{7+a_n}=\frac{(a_n+3)(2-a_n)}{7+a_n}>0$, as long as $0\leq a_n\leq2$.

Therefore $a_{n+1}\geq a_n$.

• Cleaner than what I had wrote. +1 :) Also, I had a small hole in my writing now that I think about it... – Dair Mar 28 '15 at 6:14
• I'm having a hard time coming to terms with "if 0 <= a_n<=2". – guest Mar 28 '15 at 6:25
• @guest: If you have a hard time with that, then you should go and relearn induction. Nathanson's proof above shows that we can do the following. First $a_1 \in [0,2]$. Then since we can prove that $a_2 \in [0,2]$ if we are given $a_1 \in [0,2]$, we can conclude that indeed $a_2 \in [0,2]$. But we also can prove that $a_3 \in [0,2]$ if we are given $a_2 \in [0,2]$. So we can conclude that $a_3 \in [0,2]$. Clearly we can repeat this any number of times in principle, that is, if you tell me a fixed natural number $n$, I can repeat the above argument $n-1$ times to prove that $a_n \in [0,2]$. – user21820 Mar 29 '15 at 7:27
• @guest: But this reasoning depends on one crucial assumption, that we can actually repeat the above argument as many times as we want. It for example ignores the fact that we will run out of breath/paper/life. But if we do ignore that, then we can claim that by an axiom called the axiom of induction that $a_n \in [0,2]$ for every natural number $n$, since we have at least shown that we have a way to produce a proof for any particular $n$ one may desire. – user21820 Mar 29 '15 at 7:31
• @guest: One more thing, note that you should not be having a hard time with that particular line regardless of whether you understand induction, because Nathanson is stating something that is simply true, namely that if it is true that $a_n \in [0,2]$, **then it is also true that $a_{n+1} \in [0,2]$. He proved this conditional assertion correct, and it has nothing to do with whether $a_n \in [0,2]$ is really true or not. – user21820 Mar 29 '15 at 7:39

To prove the sequence is increasing consider the function

$$f(x) = \frac{6(1+x)}{7+x}$$

and prove $f'(x)>0$ for $x>0$. To find the limit assume the limit is $a=\lim_{n\to \infty} a_n$ which gives

$$a = \frac{6(1+a)}{7+a}$$

and solve for $a$.

I leave it for you to prove boundedness.

• What's the down vote for? – science Mar 28 '15 at 5:52
• If $g(x)=x/2$, then $g'(x)=1/2>0$, but $a_1=1$, $a_{n+1}=g(a_n)=a_n/2$ is a decreasing sequence. – Nathanson Mar 28 '15 at 5:53
• Your answer is simply incorrect, as Nathanson points out a counterexample. I didn't rush and downvote, I downvoted you because your answer is plainly wrong. – user223391 Mar 28 '15 at 5:57
• @science: A function $f$ can be increasing in its domain of definition, but yet fail to satisfy the inequality $f(x)>x$ everywhere. Leaving it to you as an exercise to find examples of such functions. Anyway, with recursive definition $a_{n+1}=f(a_n)$ you need $f(x)>x$ to conclude that the sequence is increasing. So I join in your critics. – Jyrki Lahtonen Mar 28 '15 at 6:05
• You need to properly read what is written in your link. The condition is written for $a_n=f(n)$, not for $a_{n+1}=f(a_n)$. – Nathanson Mar 28 '15 at 6:46