
Re: Induction proof
Posted:
Aug 18, 2006 11:46 AM


>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.

