Search All of the Math Forum:

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

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Re: Induction proof
Replies: 1   Last Post: Aug 18, 2006 12:41 PM

 Search Thread: Advanced Search

 Messages: [ Previous | Next ]
 emailtgs@gmail.com Posts: 11 Registered: 8/17/06
Re: Induction proof
Posted: Aug 18, 2006 12:41 PM
 Plain Text Reply

okay, it's not exactly what I was looking for but I understand now what
you were implying. Thank you for the help.

kp wrote:
> He proved by induction that
> 1/2^2 + 1/3^2 + ... + 1/n^2 < 1 - 1/n
> which implies
> 1/2^2 + 1/3^2 + ... + 1/n^2 < 1 for all n >=2
>
> I think that is the utmost perfect solution you are looking for.
>
> <emailtgs@gmail.com> wrote in message
> news:1155914526.619423.102050@b28g2000cwb.googlegroups.com...

> > 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.

> >

Date Subject Author
8/18/06 Guest
8/18/06 emailtgs@gmail.com

© The Math Forum at NCTM 1994-2018. All Rights Reserved.