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Topic: Induction proof
Replies: 24   Last Post: Aug 22, 2006 4:11 PM

 Messages: [ Previous | Next ]
 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
> 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.

>

Date Subject Author
8/17/06 emailtgs@gmail.com
8/17/06 Lynn Kurtz
8/17/06 Lynn Kurtz
8/17/06 emailtgs@gmail.com
8/17/06 emailtgs@gmail.com
8/17/06 Ben Young
8/17/06 emailtgs@gmail.com
8/18/06 Paul Sperry
8/18/06 emailtgs@gmail.com
8/20/06 Brian M. Scott
8/20/06 emailtgs@gmail.com
8/20/06 Brian M. Scott
8/18/06 Torsten Hennig
8/18/06 emailtgs@gmail.com
8/18/06 Torsten Hennig
8/19/06 emailtgs@gmail.com
8/19/06 mon
8/19/06 Dave L. Renfro
8/20/06 mon
8/20/06 emailtgs@gmail.com
8/20/06 Brian M. Scott
8/19/06 Alexander Bogomolny
8/19/06 Alexander Bogomolny
8/19/06 Dave L. Renfro
8/22/06 Alexander Bogomolny