Series convergence and its nature 
Problem Statement:-
Let $\{a_n\}_{n≥1}$ be a sequence of real numbers defined recursively by
  $$a_{n+1}=\dfrac{3a_{n}}{2+a_n},$$
  for all $n ≥ 1$.
(i) If $0\lt a_1\lt1$, then prove that $\{a_n\}_{n≥1}$ is increasing and $\displaystyle\lim_{n\to\infty} {a_n} = 1$.
(ii) If $a_1\gt 1$, then prove that $\{a_n\}_{n≥1}$ is decreasing and $\displaystyle\lim_{n\to\infty} a_n = 1$.

Source-ISI-2016 UGB

My Attempt:-
I have no prior knowledge(or rather experience) of how to go about these questions, my attempt is purely based on what I could come up with and searching the mathstack site here and there.
As any noob(or is it just me) would go about these problems I started by finding $a_n$ in terms of $a_1$ which on writing a few terms(to be precise the first three terms following $a_1$) it can be seen that the 
$$a_n=\dfrac{3^{n-1}a_1}{2^{n-1}+a_{1}(3^{n-1}-2^{n-1})}$$
We can say that $$0\lt {a_n}\lt1$$
As for to where $a_n$ converges as $n\to\infty$, for that I did the follows,
$$a_n=\dfrac{a_1}{\left(\dfrac{2}{3}\right)^{n-1}+a_{1}\left(1-\left(\dfrac{2}{3}\right)^{n-1}\right)}$$
So, it becomes pretty easy to see that as $n\to\infty$, $a_n\to1$
As, for testing the nature of the series $\{a_n\}$, we can see that as $n$ increases to $\infty$ the denominator goes from $1$ to $a_1$, and since $a_1$ is smaller than $1$ so the terms of the series are increasing. But just to see that the uniform increase is followed for other $n's$(I dont know hoe to put it any other way) I tested it for $n=2$ to $n=\infty$ we see that for $a_2$ the denominator is $\dfrac{2+a}{3}$. At this point I am a little confused whether the denominator increases as we go from $n=2$ to $n=\infty$.
As for the second part of the question it is pretty easy to see that the nature of the series is increasing because as $n$ increases $a_1$ becomes more and more dominant in the denominator and as $a_1$ is greater than $1$ so the series is definitely increases and we have already proved that the series converges to $1$ as $n\to\infty$.

Please tell me how to do a more formal proof of the above question and tell me if my proof is acceptable or not(though it is incomplete).
 A: Formally finding the general term:
$$a_{n+1}=\frac{3a_n}{2+a_n}\to \frac{1}{a_{n+1}}=\frac{1}{3}\left(1+\frac{2}{a_n}\right)$$
If we call $b_n=\frac{1}{a_n}$ then
$$b_{n+1}=\frac{1}{3}\left(1+2b_n\right)\\
b_n=\frac{1}{3}\left(1+2b_{n-1}\right)\\
b_{n+1}-b_n=\frac{2}{3}(b_n-b_{n-1})$$
so the sequence $c_n=b_n-b_{n-1}$ is a GP with ratio $2/3$.
$$c_n=\left(\frac{2}{3}\right)^{n-2}c_2$$
so, applying telescopic sum we have:
$$\sum_{i=2}^{n}c_i=\sum_{i=2}^{n}(b_i-b_{i-1})=b_n-b_1$$
on the other hand
$$\sum_{i=2}^{n}c_i=\sum_{i=2}^{n}\left(\frac{2}{3}\right)^{i-2}c_2=\frac{1-\left(\frac{2}{3}\right)^{n-2}}{1-\frac{2}{3}}c_2=3c_2\left[1-\left(\frac{2}{3}\right)^{n-2}\right]$$
so,
$$b_n-b_1=3(b_2-b_1)\left[1-\left(\frac{2}{3}\right)^{n-2}\right]\\
b_n=b_1+3\left(\frac{1}{3}+\frac{2}{3}b_1-b_1\right)\left[1-\left(\frac{2}{3}\right)^{n-2}\right]\\
b_n=1+(b_1-1)\left(\frac{2}{3}\right)^{n-2}$$
So, if $a_1>1$ then $0<b_1<1$ then $b_n$ is increasing and $a_n$ is decreasing.
If $0<a_1<1$ then $b_1>1$ then $b_n$ is decreasing and $a_n$ is increasing.
In any case, $b_n\to 1$ and then $a_n\to 1$.
A: In order to prove that the sequence increases for $0<a_1<1$, I would first prove that it is bounded, namely that $a_n<1$ for every $n\in\mathbb{N}$. We can prove this by induction.
Induction basis: $a_2=\frac{3a_1}{2+a_1}<\frac{3a_1}{2a_1+a_1}=1$
Induction hypothesis: $a_k<1$ for some $k\in\mathbb{N}$
Induction step: $a_{k+1}=\frac{3a_k}{2+a_k}<\frac{3a_k}{2a_k+a_k}=1$. The induction hypothesis is needed for the inequality sign in the middle to be true.
Using this, we can prove that the sequence increases:
$$a_{n+1}=\frac{3a_n}{2+a_n}>\frac{3a_n}{2+1}=a_n$$
We used the statement proven before for the inequality sign in the middle. Note that this is all based on the fact that $0<a_1<1$ because it is used in our induction basis. Having proven that the sequence is bounded from above and increases, we have also proven that the limit of the sequence exists.
Analogously, we can prove that the sequence decreases for $a_1>1$.
