Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Math Topics » alt.math.undergrad.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
emailtgs@gmail.com

Posts: 11
Registered: 8/17/06
Re: Induction proof
Posted: Aug 19, 2006 12:52 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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.





Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.