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.

For your induction step, all you need is
$$
2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1}.
$$
That's equivalent to
$$
\frac{1}{k}-\frac{1}{(k+1)^2}-\frac{1}{k+1}\geq 0
$$
i.e.
$$
\frac{(k+1)^2-k-k(k+1)}{k(k+1)^2}=\frac{1}{k(k+1)^2}\geq 0.
$$
So it holds.
