Date: Aug 19, 2006 1:08 PM
Author: mon
Subject: Re: Induction proof
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

news:1156006369.232048.70870@i3g2000cwc.googlegroups.com...

> 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.

>