# Prove that $a_{n+1}=3\frac{a_{n}+1}{a_{n}+3} , a_{1} =a> 0$ is convergent

The sequence $a_{n}$ is given in this recursive form:

$$a_{n+1}=3\frac{a_{n}+1}{a_{n}+3} , a_{1} =a> 0$$

How does one show that it is convergent?

I tried to prove that it is monotone increasing and bounded by $\sqrt3$ but than I thought that it may be better to seperate into cases which $a>\sqrt3$ and $a<\sqrt3$,

Either way I did'nt get to a solution and wasn't able to write the proper Induction.

Thank you

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Don't put math operators outside the dollar signs; especially not $<$ and $>$, since they get interpreted as HTML tag delimiters. – joriki Apr 21 '11 at 12:49
Is it convergent? Doesn't $\lim_{n \rightarrow \infty} a_n= 3$ by L'Hopital's? – Covi Apr 21 '11 at 12:56
@Covi: No, L'Hôpital applies neither to recursion relations for sequences, nor to quotients where the numerator and denominator don't vanish. – joriki Apr 21 '11 at 13:06
Maybe in this form it's easier: $a_{n+1} = 3-\frac{6}{a_n +3}$. I think if $a_1<\sqrt{3} \quad$ $a_n$ is monotone increasing and upper bounded by 3, and if $a_1>\sqrt{3} \quad$ $a_n$ is monotone decreasing and lower bounded by 0. – tpv Apr 21 '11 at 13:14
Abstract duplicate of math.stackexchange.com/questions/7266/… . – Qiaochu Yuan Apr 21 '11 at 18:22

I will sketch half of a solution of the problem. You can do the second half.

If $a$ is positive, all terms are obviously positive. Suppose that $a<\sqrt{3}$. We show that the sequence is bounded and increasing.

First show that the sequence is bounded. We show that if $a_n^2<3$, then $a_{n+1}^2 <3$.

Calculate: $$3-a_{n+1}^2=3-\frac{9(a_n+1)^2}{(a_n+3)^2}=\frac{6(3-a_n^2)}{(a_n+3)^2}$$ so if $a_n^2<3$ then $a_{n+1}^2<3$.

Now show that the sequence is increasing. Note that $$a_{n+1}-a_n=\frac{3(a_n+1)}{a_n+3}-a_n=\frac{3-a_n^2}{a_n+3}$$ By what was shown eaarlier, the right-hand side of the above equation is positive.

Now it's your turn. Suppose that $a>\sqrt{3}$. Continue.

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You are on he right track. Let $f(x)=3(x+1)/(x+3)$. It is a continuous, increasing function. It cuts the diagonal of the first quadrant (the line $y=x$) at only one point: $x=\sqrt3$, $y=\sqrt3$. If $0\le x<\sqrt3$, then $x<f(x)<\sqrt3)$; this is all you need to prove (by induction) that in this case $\{a_n\}$ is a bounded increasing sequence. On the other hand, if $x>\sqrt3$, then $\sqrt3<f(x)<x$; this implies that $\{a_n\}$ is decreasing and bounded below by $\sqrt3$. Once you know the sequence is convergent, taking limits on both sides of the recurrence relation shows that the limit must be a solution of the equation $f(x)=x$. Since the only positive solution is $x=\sqrt3$, this is the limit.

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You are correct to divide the cases (there is also $a=\sqrt{3}$)

• For $a_n =\sqrt{3}$, you should be able to show $a_{n+1} =\sqrt{3}$
• For $0 < a_n < \sqrt{3}$, you should be able to show $a_n < a_{n+1} < \sqrt{3}$
• For $a_n > \sqrt{3}$, you should be able to show $a_n > a_{n+1} > \sqrt{3}$

So you either have a constant, or an increasing sequence bounded above or a decreasing sequence bounded below. In any case you have a converging sequence.

If it helps, you have $a_{n+1}=3 - \dfrac{6}{a_{n}+3}$

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If the sequence has a limit $\alpha$ then $\alpha$ has to satisfy the equation $\alpha=3{\alpha+1\over\alpha+3}$, so $\alpha$ would have to be one of $\pm\sqrt3$, and as all $a_n$ are $> 0$, we are left with $\alpha=\sqrt3$. This suggests putting $a_n=\sqrt3+b_n$ $\ (n\geq 1)$ in the hope that we can prove $\lim_{n\to\infty} b_n=0$. From $$\sqrt3 + b_{n+1}=3{\sqrt3 + b_n+1 \over \sqrt3 + b_n+3}$$ we get the following recursion formula for the $b_n$: $$b_{n+1}=b_n \ {3-\sqrt3 \over \sqrt3 + b_n+3}= b_n\ {3-\sqrt3 \over a_n+3}\ .$$ Here the last fraction is positive and $< (3-\sqrt3)/3<{1\over2}$ from which it follows that the $b_n$ indeed converge to $0$.

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A pedagogically very good approach! The main difficulty for students new to these calculations is "How do I start?" Their intuitions is likely better for convergence to $0$, so looking at the difference is excellent heuristic advice. – André Nicolas Apr 21 '11 at 14:51