Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume 
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$
This is where I get stuck. Any help appreciated.

$\displaystyle \frac{1}{i^2}< \frac{1}{i(i-1)}=\frac{1}{i-1}-\frac{1}{i}$,for $i\ge 2$
So we have $\displaystyle 1+\sum_{i=2}^{k}\frac{1}{i^2}\le 1+\sum_{i=2}^{k}(\frac{1}{i-1}-\frac{1}{i})=2-\frac{1}{k}$
I think this is better than induction.
