Date: Aug 19, 2006 12:52 PM
Author: emailtgs@gmail.com
Subject: Re: Induction proof
okay, my question is, just how do you decide to go from here ... to

here.

< (1 - 1/n) + 1/(n+1)^2 (by induction hypotheses)

< (1 - 1/n) + 1/(n*(n+1))

I know you've made a note of what you did but I don't understand how

you can do that.

Would you please explain. Thank you

Torsten Hennig wrote:

> >Hi Torsten, thank you for your reply however you're >solving a different

> >problem here. It appears that you've introduced a -1/n >to the right of

> >the inequality for your convinience. That isn't the >orignial problem.

> >I'm not sure what you're doing at all. Please everyone, >here's the

> >problem ((((( 1/2^2 + 1/3^2 + ... + 1/n^2 < 1 ))))) FOR >ALL n,

> >greater than or equal to 2, PROOF by INDUCTION. I only >capitalized for

> >clarity, not yelling here.

> >

> >Torsten Hennig wrote:

> >> >Prove by induction that 1/2^2 + 1/3^2 + ... + 1/n^2 < >1 >Please help!

> >> >Thank you!

> >>

> >> Hi,

> >>

> >> show by induction that

> >> 1/2^2 + 1/3^2 + ... + 1/n^2 < 1 - 1/n

> >> In the induction step, use that

> >> 1/(n+1)^2 < 1/(n*(n+1)) = 1/n - 1/(n+1) .

> >>

> >> Best wishes

> >> Torsten.

>

> Hi,

>

> you want to show that

> 1/2^2 + 1/3^2 + ... + 1/n^2 < 1

> for all n >= 2.

>

> If you can show (e.g. by induction) that

> 1/2^2 + 1/3^2 + ... + 1/n^2 < 1 - 1/n for all n >= 2,

> you have what you want because 1 - 1/n < 1.

>

> Induction start : 1/2^2 = 1/4 < 1/2 = 1 - 1/2 ( < 1 )

> Induction step :

> (1/2^2 + 1/3^2 + ... + 1/n^2) + 1/(n+1)^2

> < (1 - 1/n) + 1/(n+1)^2 (by induction hypotheses)

> < (1 - 1/n) + 1/(n*(n+1))

> = (1 - 1/n) + (1/n - 1/(n+1))

> = 1 - 1/(n+1)

> ( < 1 )

>

> Best wishes

> Torsten.