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.