Prove $1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}$ by Induction The Question
Prove $1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}$ where $n\ge2$ and $n$ is an integer by Induction
My Work
Basis Step:
1 + $\frac{1}{4} = \frac{5}{4}$
$2-\frac{1}{2} = \frac{3}{2}$
$\frac{5}{4}<\frac{3}{2}$
Inductive Hypothesis:
$1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} < 2 - \frac{1}{k}$
Induction Step
We must show $1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} + \frac{1}{(k+1)^2} < 2 - \frac{1}{(k+1)}$
$1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} < 2 - \frac{1}{k}$
$2 - \frac{1}{k} + \frac{1}{(k+1)^2} = 2 - \frac{(k^2+2k+1)+k}{k(k+1)^2} = 2-\frac{k^2 + 3k + 1}{k^3 +2k^2+k}$
My Problem
I can't seem to complete the proof. Any hints on how to change my approach for success?

$\displaystyle 2-\frac{k^2+3k+1}{k^3+2k^2+k}=$
$\displaystyle 2-\frac{k+3+1/k}{k^2+2k+1}=$
$\displaystyle 2-\frac{k+3+1/k}{(k+1)^2}=$
$\displaystyle 2-\frac{k+1+2+1/k}{(k+1)^2}=$
$\displaystyle =2-\frac{1}{k+1}-\frac{2+1/k}{(k+1)^2}<$
$\displaystyle 2-\frac{1}{k+1}$
