Find the limit of recursive sequence, if it exists: $a_{n+1}=\frac{7+3a_n}{3+a_n}$

Question

My goal is to to test this recursive sequence if it's convergent and if yes, find the limit.

$$a_1=3,\:a_{n+1}=\frac{7+3a_n}{3+a_n}$$

I know how to do this with normal sequences, but this is the first time we have to use a recursive sequence.

If you calculate the first few parts you get:

n | a(n)
1 | 3
2 | 2.66667
3 | 2.64706
4 | 2.64583
5 | 2.64576
6 | 2.64575
7 | 2.64575
8 | 2.64575

So I take it the sequence is convergent against 2.64575... But how do you prove that in a calculation?

Answer 1

Rewrite the recursion, for example

$$a_{n+1}=\frac{7+3a_n}{3+a_n}=\frac{9+3a_n}{3+a_n}-\frac{2}{3+a_n}=3-\frac{2}{3+a_n} $$ First of all, if $a_n>0$, then $a_{n+1}>3-\frac23>0$ as well. Hence by inductiuon (and as $a_1>0$), $a_n>0$ for all $n$. This shows that $\{a_n\}$ is bounded from below, as suspected.

Next observe that the bigger (positive) $x$ is, the smaller is $\frac2{3+x}$ and the bigger is $4-\frac2{3+x}$. Consequently, if $a_n>a_{n+1}$, then also $a_{n+2}=3-\frac2{3+a_{n+1}}>3-\frac2{3+a_{n}}=a_{n+1}$. Again, using the first instance of this, i.e. $a_1=3>a_2=\frac83$, we conclude by induction that $\{a_n\}$ is (strictly) decreasing.

From the above we se that the limit $a:=\lim_{n\to\infty}a_n$ exists (why?). Then also $$a =\lim_{n\to\infty}a_{n+1}= \lim_{n\to\infty}\frac{7+3a_n}{3+a_n}=\frac{7+3\lim_{n\to\infty}a_{n}}{3+\lim_{n\to\infty}a_{n}}=\frac{7+3a}{3+a}.$$ Can you find from this, which values of $a$ are candidates for the limit? Which of these cannot be the limit? Hence ... ?

Answer 2

If you assume that it exists, one says, $a_n \to L$, then $a_{n+1} \to L$ since $(a_{n+1})$ is a subsequence of $(a_n)$ and then, it must converge to the same limit. Now, if you apply the limit as $n \to \infty$ in the equality that definines recusively the sequence, you find:$$L = \frac{7+3L}{3+L}.$$ Solvind this equation for $L$, you find $L= \pm \sqrt{7}$. Since every term in the sequence is positive, the limit cannot be negative. Then, $L= \sqrt{7}$.