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.

Let's assume it's true for $k$.
Then $\sum_{i=1}^k\dfrac{1}{i^2}\leq2-\dfrac 1 k$
Let's try $k+1$.
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}=\sum_{i=1}^k\dfrac{1}{i^2}+\dfrac{1}{(k+1)^2}\leq2-\dfrac 1 k+\dfrac{1}{(k+1)^2}$
As $\dfrac{1}{(k+1)^2}-\dfrac 1 k<-\dfrac{1}{k+1}$ ,
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}\leq2-\dfrac{1}{(k+1)^2}$
which shows that it is true for $k+1$.
