<s>[INST] 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?
 [/INST] You need to show that
$$2-{1\over k}+{1\over(k+1)^2}\le2-{1\over k+1}$$
This is equivalent to showing
$${1\over k+1}+{1\over(k+1)^2}\le{1\over k}$$
and, by clearing out the denominators, this is equivalent to showing
$$k((k+1)+1)\le(k+1)^2$$
Can you take it from here?
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