Prove $1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}$ by Induction

Question

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?

Answer 1

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?

Answer 2

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