(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

prove by induction that

[tex]\frac{1}{n^2} < \frac{1}{n(n-1)}[/tex]

2. Relevant equations

n/a

3. The attempt at a solution

Let

[tex]P(n):\frac{1}{n^2} < \frac{1}{n(n-1)}[/tex]

Since P(1) is undefined, the base case is P(2)

[tex]P(2) = \frac{1}{2^2} < \frac{1}{2(2-1)}[/tex]

[tex]P(2) = \frac{1}{4} < \frac{1}{2}[/tex]

Therfore P(n) is true for all n>1

Inductive Step

Assume [tex]P(k)[/tex] is true.

[tex]P(k):\frac{1}{k^2} < \frac{1}{k(k-1)}[/tex]

I know what I need my inequality to look like for [tex]k+1[/tex] I am just having problems getting there. my [tex]k+1[/tex] will look like:

[tex]P(k+1):\frac{1}{(k+1)^2} < \frac{1}{(k+1)[(k+1)-1]}[/tex]

after simplification:

[tex]P(k+1):\frac{1}{(k+1)^2} < \frac{1}{(k+1)(k)}[/tex]

So I know where I need to be. Consider this is just side work.

So for [tex]k+1[/tex] I can rewrite an expansion on the LHS as:

[tex]\frac{1}{k^2+2k+1}[/tex]

And based on my inductive step the RHS of the inequality is:

[tex]\frac{1}{[k(k-1)]+2k+1}[/tex]

And this is where I am hitting the block. Try as I might, I can't see any steps to take to make [tex]\frac{1}{[k(k-1)]+2k+1}[/tex] equivalent to [tex]\frac{1}{(k+1)(k)}[/tex]

I am worried that I may have set up my induction incorrectly though.

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# Proof by induction

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