
Re: Induction proof
Posted:
Aug 19, 2006 12:52 PM


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.

