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?

Your base case is wrong. You should realize it's true since $\frac{5}{4}<\frac{3}{2}$ obtained from $$1+\frac{1}{2^2} = \frac{5}{4} < \frac{3}{2} = 2-\frac{1}{2}$$ For the induction step, suppose $P(n)$ is true for all $n \in \{1,2,\ldots,k\}$. Then $$\begin{align}1+\frac{1}{2^2}+\ldots+\frac{1}{k^2}+\frac{1}{(k+1)^2} = \left(1+\frac{1}{2^2}+\ldots+\frac{1}{k^2}\right)+\frac{1}{(k+1)^2} \\ < \left(2-\frac{1}{k} \right)+\frac{1}{(k+1)^2}\end{align}$$ Now if you can show that $$\left(2-\frac{1}{k} \right)+\frac{1}{(k+1)^2}<2-\frac{1}{k+1}$$ you are done. It is possible to get to that inequality simply starting with the fact that $\frac{1}{k+1}<\frac{1}{k}$
