Convergence of Sequence of Real Numbers

Define a sequence of real numbers recursively as follows. Let $a_1 = 1$ and $a_{n+1} = 1 + \frac{1}{1+a_n}$.

First, show the sequence is not monotonic.

Second, show that $a_n \geq 1$ for all $n$ and conclude that the sequence is super Cauchy.

First, observe that if $a_n < \sqrt{2}$, then $a_{n+1} > 1 + \frac{1}{1+\sqrt{2}} = 1 + \sqrt{2} - 1 = \sqrt{2}$, and similarly if $a_n > \sqrt{2}$ then $a_{n+1} < \sqrt{2}$. So in fact $a_n$ cannot be eventually monotonic as it alternates between being greater and less than $\sqrt{2}$.

As mentioned in the comments, one can see by induction that $a_n \geq 1$ for all $n$, where in the inductive step we use only that $a_n$ is positive.

Finally, we have \begin{eqnarray}|a_{n+1} - a_n| &=& |1 + \frac{1}{1+a_n} - a_n| = |\frac{1}{1+a_n} - \frac{1}{1+a_{n-1}}|\\ &=& |\frac{a_{n-1}-a_n}{(1+a_{n-1})(1+a_n)}| < \frac{1}{4}|a_n-a_{n-1}|,\end{eqnarray} where we used $a_n, a_{n-1} \geq 1$ in the last step. So $a_n$ is super Cauchy.

One can conclude from the first observation together with the fact that the sequence is Cauchy that $\displaystyle \lim_{n\rightarrow \infty} a_n = \sqrt{2}$.