Jtdylktuy

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:

$$a_n-a_{n+1}>0Rightarrow a_n-{frac{3(1+a_n)}{3+a_n}}>0Rightarrow a_n^2>3Rightarrow a_n>sqrt{3}$$ 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?

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>0implies a_{n+1}>0 $ for any $n$. So by induction we have $a_n>0$ for all $nin Bbb N.$

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

Since $3+a_n>a_n>0$ we have $$a_n>a_{n+1} iff (3+a_n)left(a_n-frac {3(1+a_n)}{3+a_n}right)>0 iff$$ $$iff 3a_n+a_n^2> 3+3a_niff a_n^2>3,$$ 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 $Lgeq 0$ . And we have $$L=lim_{nto infty}a_{n+1}=lim_{nto infty}frac {3(1+a_n)}{3+a_n}= frac {3(1+L)}{3+L}.$$ So $L=frac {3(1+L)}{3+L}, $
which implies $L^2=3.$ Now $(Lgeq 0land L^2=3)implies L=sqrt 3.$