Prove $\sum_1^n 1/i^2 \le 2 - 1/n$ for all natural $n$. 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?

$$
{1\over(k+1)^2}<{1\over k(k+1)}={k+1-k\over k(k+1)}=\frac1k-{1\over k+1}
$$
As a result, we have
$$
\sum_{i=1}^{k+1}{1\over i^2}\le2-\frac1k+\frac1k-{1\over k+1}=2-{1\over k+1}
$$
Hence, by the principle of mathematical induction, we conclude that the proposition holds for all positive integer $n$.
