Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality 

By induction assume it is true for $n$.
Then, by induction,
$\sum_{k = 1}^{n+1} \frac{1}{k^{2}} =\sum_{k = 1}^{n} \frac{1}{k^{2}} +\frac{1}{(n+1)^2} < 2-\frac{1}{n} +\frac{1}{(n+1)^2}$.
So to finish we need $ 2-\frac{1}{n} +\frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}$,
or $\frac{1}{n+1} \leq \frac{1}{n}-\frac{1}{(n+1)^2}$, which is true for $n>1$. 
