Establish the convergence and find the limit of $a_n$

Question

I've just been introduced to the idea of series and I'm currently stuck on this problem:

Establish the convergence and find the limit of $a_n$ if $$a_1 = 1, a_{n+1} = \frac{2(2a_n + 1)}{a_n+3}, n \in \mathbb N$$

I figured that I'm suppose the prove that the series is increasing and bounded but I'm not sure how to prove either.

Answer 1

Note that all terms will be positive. A limit $l$ would have to satisfy $l = \frac{2(2l + 1)}{l+3}$ which has positive solution $l=2$. We can use that to find an easy solution by considering the series of terms $a_n - 2$ and proving this series tends to $0$.

We have $ a_{n+1}-2 = \frac{2}{a_n+3}(a_n - 2) $. Now $\frac{2}{3}>\frac{2}{a_n+3}>0$ and so $ |a_{n+1}-2|< \frac{2}{3}|a_n - 2| $. The required result is now obvious with $|a_2 - 2| $ less than $\frac{2}{3}$, $|a_3 - 2| $ less than $\frac{4}{9}$, etc.

Answer 2

Note that $$a_{n+1} = \frac{2(2a_n + 1)}{a_n+3} =4-\frac {10}{a_n+3}$$

Thus the sequence is increasing and bounded by $4$ thus it is convergent with a limit of $l$

The limit satisfies $$l=\frac {4l+2}{l+3}$$

Solving for $l$ and choosing the positive answer results in $l=2$