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

Is it okay to (unrigorously) state that

converges because does?Thanks.

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

Called a comparison test? http://en.wikipedia.org/wiki/Comparison_test

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

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

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

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

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

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

Ricky wrote:

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

isrigorous.

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

Daniel123 wrote:

Ricky wrote:Just wanted to note that stating the sum converges because 1/n^2 does

isrigorous.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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