sequences and series - How to find the monotony and the convergence of this function?

I have to find the monotony and the convergence of

$$a_{n+1} = 2\frac{1+a_{n}}{3+a_{n}}$$
for every $n \in\mathbb N$ when $0

What I have done is:

I said that if $\lim_{n\to\infty} a_n$ exists and is equal to a number $l$, then

$$\begin{align*} l&=\lim_{n\to\infty} a_{n+1}\\ &=\lim_{n\to\infty}\left(2\frac{1+a_{n}}{3+a_{n}}\right)\\ &=2\cdot\frac{1+\lim_{n\to\infty} a_{n}}{3+\lim_{n\to\infty} a_{n}}\\ &=2\cdot\frac{1+l}{3+l}\\ \Leftrightarrow l & = 2\cdot\frac{1+l}{3+l}\\ \Leftrightarrow 3l+l^2 &=2+2l\\ \Leftrightarrow l^2+l-2 &=0 \end{align*}$$

And from here I find that $l=1$ and $l=-2$. But, since $0

From here on I don't know how to continue. I have solved a couple of similar exercises, but on them I was given a number for $a_{1}$. So I used inductive reasoning (not sure if this is the right expression) and I found the monotony of the function.

The way I used the inductive reasoning is:

• I saw that for $n=1$ the statement was true



• I supposed it was true for $n$ and then proved that it was true for $n+1$.

But since I don't have a number for $a_{1}$ I can't continue.

Any tips?

Answer

I solved it after all. I took three cases for the $a_{n}$.

The cases where:

• $a_{1}<1$


• $a_{1}=1$


• $a_{1}>1$

For $a_{1}<1$:

I found (per @Bill Province's comment) that the function is monotonically increasing and that $a_{k}<1$. Thus it is convergent.

For $a_{1}=1$:

I found that $a_{k}=1$, thus it is convergent.

For $a_{1}>1$:

I found that $a_{k}>1$. Τhus the $(a_{n})$ has a lower bound; it also is monotonically increasing -> thus is is not convergent.