|
|
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.
|
|
