I'm trying to do this by induction. It works for $n=1$ because we get $1 \le 2 - 1 = 1$. Now suppose for some natural $k \ge 1$, we have $\sum\_{i=1}^k \le 2 - 1/k$. I must show $\sum\_{i=1}^{k+1} \le 2 - 1/(k+1)$. I started out from the hypothesis and added $1/(k+1)^2$ to both sides of the inequality. I get

\begin{align\*}
 1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} + \frac{1}{(k+1)^2} \le 2 - \frac{1}{k} + \frac{1}{(k+1)^2}
\end{align\*}

and it's not obvious what to do from there. Any hints?