The sequence is defined as such, with $a_1=1$, $$ a_{n+1} = \begin{cases} a_n + 1/n, & \mbox{if } a_n^2 \leq 2 \\ a_n - 1/n, & \mbox{if } a_n^2 > 2 \\ \end{cases}. $$
In the book, P.M. Fitzpatrick's Advanced Calculus, the exercises wants me to show that for all indices $n$, $|a_n-\sqrt{2}|<2/n$. Of course after that it's easy to prove that the sequence converges by the Comparison Lemma, since $\{ 1/n\}$ converges. My problem is with proving that inequality. The only way I can think of is by induction. This is my incomplete attempt:
Since $1<\sqrt{2}<2$, we have $-1<1-\sqrt{2}<0$ and so $|a_1 - \sqrt{2}|<1<2$.
Suppose that $|a_k-\sqrt{2}|<2/k$. If $a_k^2 \leq 2$ then $a_{k+1}=a_k+1/k$.
Now this is where I get stuck. Of course I would also have to do it for the case when $a_k^2>2$, but that comes after. I have tried different ways of manipulating the last two inequalities but to no avail. Any suggestions?