Real Analysis problem concerning recursive sequence limit. Define $$a_1=1$$, and define
$$
a_{n+1} = \left\{\begin{aligned}
&a_n+1/n &&: a_n^2\leq 2\\
&a_n-1/n &&: a_n^2>2
\end{aligned}
\right.$$
I am then asked to show that $$|a_n-\sqrt{2}|<2/n$$ for all indices n in order to show convergence by comparison.
My initial thought was to use induction but my algebra during the inductive step doesn't go very far. Any thoughts would be greatly appreciated.

Induction works. Suppose $|a_n - \sqrt 2| < 2/n$, i.e.
\begin{equation*}
(1) \quad \sqrt 2 - 2/n < a_n < \sqrt 2 + 2/n.
\end{equation*}
There are two cases:
Case 1. If $a_n^2 \le 2$, i.e. $a_n \le \sqrt 2$, we can refine (1) to
\begin{equation*}
\sqrt 2 - 2/n < a_n \le \sqrt 2.
\end{equation*}
And we have $a_{n + 1} = a_n + 1/n$, so by adding $1/n$ to this inequality, we get
\begin{equation*}
\sqrt 2 - 1/n < a_{n + 1} \le \sqrt 2 + 1/n,
\end{equation*}
i.e. $|a_{n + 1} - \sqrt 2| < 1/n$.
Case 2. If $a_n^2 \ge 2$, i.e. $a_n \ge \sqrt 2$, we can refine (1) to
\begin{equation*}
\sqrt 2 < a_n \le \sqrt 2 + 2/n.
\end{equation*}
And we have $a_{n + 1} = a_n - 1/n$, so by subtracting $1/n$ from this inequality, we get
\begin{equation*}
\sqrt 2 - 1/n < a_n \le \sqrt 2 + 1/n,
\end{equation*}
i.e. $|a_{n + 1} - \sqrt 2| < 1/n$.
Either way, we have $|a_{n + 1} - \sqrt 2| < 1/n$. And for all $n \ge 1$, we have $1/n \le 2/(n + 1)$.
(Proof of the last inequality:
\begin{equation*}
2/(n + 1) - 1/n = (2n - n - 1)/(n(n + 1)) = (n - 1)/(n^2 + n),
\end{equation*}
which is nonnegative).
