Proving inequality $\ 1+\frac14+\frac19+\cdots+\frac1{n^2}\le 2-\frac1n$ using induction Question:

Prove $$\ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n},
\text{ for all natural } n$$

My attempt:
Base Case: $n=1$ is true:
I.H: Suppose $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}\le 2-\frac{1}{k},$ for some natural $k.$
Now we prove true for $n = k+1$
$$ 1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2},\text{ by induction hypothesis} $$
Now how do I show that $2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{\left(k+1\right)}\text{ ?}$
Have I done everything correctly up until here? 
If yes, how do I show this inequality is true? 
Any help would be appreciated. 

You are right!
We need to prove that:
$$\frac{1}{(k+1)^2}<\frac{1}{k}-\frac{1}{k+1}$$ or
$$\frac{1}{(k+1)^2}<\frac{1}{k(k+1)},$$
which is obvious.
