limit, low or high bound, convergence for recursive sequence [duplicate]

If $a=\frac1{1+a}$, then $$\tag1a_{n+1}-a=\frac1{1+a_n}-\frac1{1+a}=\frac{a-a_n}{(1+a_n)(1+a)} $$ Now if $a>0$ and $a_n>0$, this implies that $$ |a_{n+1}-a|\le \frac1{1+a}\cdot |a_n-a|=q\cdot |a_n-a|$$ with $0<q=\frac1{1+a}<1$. Then by induction $$|a_n-a|\le q^{n-1}\cdot |a_0-a|$$ i.e. $a_n\to a$.

Remark: From $(1)$ we see that $a_n-a$ changes sign at each step, hence the sequence $(a_n)_{n\in\mathbb N}$ is indeed not monotonic. But subsequences $(a_{2n})$ and $(a_{2n+1})$ are.

From $0 \lt a_n \lt 1$ you have proven that the sequence has at least one accumulation point in the interval. What could go wrong is that the sequence could bound around the neighborhoods of several points. To show that it is a limit, write $a_n=a+e_n$, so $e_n$ is the error of term $n$. Then $e_{n+1}=a_{n+1}-a=\frac 1{1+a+e_n}-\frac 1{1+a}=\frac {-e_n}{(1+a+e_n)(1+a)}$. As long as $e_n$ is not too large and negative (I leave it to you to argue this) the series will converge to zero, and we have a limit.