mon
Posts:
46
Registered:
12/6/04


Re: Induction proof
Posted:
Aug 19, 2006 1:08 PM


I want to understand why 1/(n+1)^2 < 1/(n*(n+1))
it is because: 1/(n+1)^2 = 1/ [(n+1) * (n+1)] < 1 / [ n * (n+1)]
or another way n+1 > n (n+1)*(n+1)> n * (n+1) (multiply both sides by n+1 which is positive) (n+1)^2 > n * (n+1) 1 / (n+1)^2 < 1 / (n * (n+1)) (when we take reciprocal inequality changes direction)
<emailtgs@gmail.com> wrote in message news:1156006369.232048.70870@i3g2000cwc.googlegroups.com... > 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. >

