Proving $ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction Question:

Let $P(n)$ be the statement that $1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} <2- \dfrac{1}{n}$. Prove by mathematical induction. Use $P(2)$ for base case.

Attempt at solution:
So I plugged in $P(2)$ for the base case, providing me with $\dfrac{1}{4} < \dfrac{3}{2}$ , which is true. 
I assume $P(n)$ is true, so I need to prove $P(k) \implies P(k+1)$. 
So $\dfrac{1}{(k+1)^2} < 2 - \dfrac{1}{k+1}$.
I don't know where to go from here, do I assume that by the Inductive hypothesis that it's true?

Hint: You assume that the statement is true for $n=k$.  In other words,
$$
1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}<2-\frac{1}{k}.
$$
Now, add $\frac{1}{(k+1)^2}$ to both sides to get
$$
1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}<2-\frac{1}{k}+\frac{1}{(k+1)^2}.
$$
What you would really like is that
$$
2-\frac{1}{k}+\frac{1}{(k+1)^2}<2-\frac{1}{k+1}
$$
because then, by transitivity, your result would hold.  So, can you prove that?  
