# Show that the sequence $a_1=3$ and $a_{n+1}=\frac{3(1+a_n)}{3+a_n}$ converges to $\sqrt3$

First of all let´s see how it goes: $$3,\frac{3(1+3)}{3+3}=\frac{12}{6}=2,\frac{3(1+2)}{3+2}=\frac{9}{5}=1.8,...$$ We see that it's decreasing.

What i need to show first is that the sequence decreases. To do that, i´d say:

$$a_n-a_{n+1}>0\Rightarrow a_n-{\frac{3(1+a_n)}{3+a_n}}>0\Rightarrow a_n^2>3\Rightarrow a_n>\sqrt{3}$$ But i think this is like say that $$a_n$$ converges to $$\sqrt3$$, which is asked to be proved. So, How can i show this sequence decreases?. And to show it's bounded?

• Still study the difference $a_n -\sqrt 3$. – xbh Nov 16 '18 at 2:15

Hint. If you can prove that it has some limit $$L$$ then you can take the limit of both sides of the recursive definition and conclude ...

Hint:

• Let $$f(x)=\frac{3(1+x)}{3+x}$$. Prove that $$f(x) \le x$$ for $$x \ge \sqrt 3$$

• Prove that $$x \ge \sqrt 3 \implies f(x) \ge \sqrt 3$$

Conclude that $$a_n$$ is decreasing and bounded below and hence converges. Since $$f$$ is continuous, $$a_n$$ converges to fixed point of $$f$$.

You can use induction to show $$a_n>a_{n+1}$$: $$1) \ a_1=3>2=a_2;\\ 2) \ \text{assume} \ a_{n-1}\color{red}>a_n;\\ 3) \ a_{n}=\frac{3(a_{n-1}+1)}{3+a_{n-1}}=3-\frac{6}{3+a_{n-1}}\color{red}>3-\frac6{3+a_n}=\frac{3(a_n+1)}{3+a_n}=a_{n+1}.$$

We have $$a_1>0$$ and from the definition of $$a_{n+1}$$ we have $$a_n>0\implies a_{n+1}>0$$ for any $$n$$. So by induction we have $$a_n>0$$ for all $$n\in \Bbb N.$$

We have $$a_1>\sqrt 3$$. If $$a_n>\sqrt 3$$ then, since $$a_{n+1}>0,$$ we have $$a_{n+1}>\sqrt 3\iff a_{n+1}^2>3\iff$$ $$\iff \frac {3^2(1+a_n)^2}{(3+a_n)^2}>3\iff$$ $$\iff 3^2(1+a_n)^2 -3(3+a_n)^2>0\iff$$ $$\iff 6(a_n^2-3)>0.$$ So by induction we have $$a_n>\sqrt 3$$ for all $$n\in \Bbb N.$$

Since $$3+a_n>a_n>0$$ we have $$a_n>a_{n+1} \iff (3+a_n)\left(a_n-\frac {3(1+a_n)}{3+a_n}\right)>0 \iff$$ $$\iff 3a_n+a_n^2> 3+3a_n\iff a_n^2>3,$$ and we have already shown that $$a_n^2>3.$$ So $$a_n>a_{n+1}$$ for all $$n.$$

Since $$(a_n)_n$$ is a decreasing sequence of positive terms it has a limit $$L\geq 0$$ . And we have $$L=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\frac {3(1+a_n)}{3+a_n}= \frac {3(1+L)}{3+L}.$$ So $$L=\frac {3(1+L)}{3+L},$$ which implies $$L^2=3.$$ Now $$(L\geq 0\land L^2=3)\implies L=\sqrt 3.$$