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#1 2009-01-16 09:37:04

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Really, really quick

Is it okay to (unrigorously) state that

converges because
does?

Thanks.

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#2 2009-01-16 09:46:24

Daniel123
Member
Registered: 2007-05-23
Posts: 663

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#3 2009-01-16 09:52:54

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Really, really quick

I don't even know why I asked this, it obviously converges.

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#4 2009-01-16 12:26:03

Muggleton
Member
Registered: 2009-01-15
Posts: 65

Re: Really, really quick

There's a rather cute name for this test. It's called the Plain Vanilla Comparison Test.

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#5 2009-01-16 12:49:44

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Really, really quick

Just wanted to note that stating the sum converges because 1/n^2 does is rigorous.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#6 2009-01-17 00:41:11

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Really, really quick

Ricky wrote:

Just wanted to note that stating the sum converges because 1/n^2 does is rigorous.

Ahh right, okay. I thought I might have to write out some |a_n| < |b_n| things mentioned on that wiki page.

Thanks

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#7 2009-01-17 01:23:46

Kurre
Member
Registered: 2006-07-18
Posts: 280

Re: Really, really quick

Daniel123 wrote:
Ricky wrote:

Just wanted to note that stating the sum converges because 1/n^2 does is rigorous.

Ahh right, okay. I thought I might have to write out some |a_n| < |b_n| things mentioned on that wiki page.

Thanks

yea but obv 0<1/(1+n^2)<1/n^2 so |a_n|<|b_n| for all n. Imo you should do this calculation to show that the sum converges, not just state it without motivation.

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