proof verification - 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.

Answer

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.