Help with proof using induction: $1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ I am having trouble with the following proof:

For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$

My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong. 

Inductive Hypothesis: suppose $\sum_{k=1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$. 
Inductive Step: then suppose $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq 2 - \frac{1}{n} + \frac{1}{(1+n)^2}. $
So it suffices to show that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$.
But this, by simple algebra, is equivalent to $ n \leq n+1  $, which is obviously true $\forall n \in \mathbb{N}$.
EDIT Proof that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$ is equivalent to $n \leq n+1$. 
Add $\frac{1}{n}$ to both sides to se that what we need to show is $\frac{1}{(1+n)^2}  \leq \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}.$ Multiply both sides by $(n+1)$. Then take reciprocals and remember to swap the inequality sign.    
