```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 ... tohere.< (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 howyou can do that.Would you please explain. Thank youTorsten 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.
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