1. Mathematical Induction

Hi, I started this problem, got stuck, and need a hint to continue.

Prove that $\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n}$

Base case:
$\displaystyle \frac{1}{1^2} = 2 - 1$
$\displaystyle 1 =1$ so the inequality holds for $\displaystyle n =1$.

Induction Hypothesis:
Assume that
$\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} \leq 2 - \frac{1}{k}$, where $\displaystyle k$ is an arbitrary integer $\displaystyle \geq 1$.

Prove that $\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k+1}$

Here is what I got:
$\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k} + \frac{1}{(k+1)^2}$

Any suggestion on how to continue?
2. . . . . .$\displaystyle \frac{1}{k}\, -\, \frac{1}{(k\, +\, 1)^2}\, =\, \frac{(k\, +\, 1)^2\, -\, k}{k(k\, +\, 1)^2}$
. . . . .$\displaystyle =\, \frac{k^2\, +\, k\, +\, 1}{k(k\, +\, 1)^2}$
. . . . .$\displaystyle >\, \frac{k^2\, +\, k}{k(k\, +\, 1)^2}\, =\, \frac{k(k\, +\, 1)}{k(k\, +\, 1)^2}$
. . . . .$\displaystyle =\, \frac{1}{k\, +\, 1}$