Math Is Fun Forum

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Really, really quick

Is it okay to (unrigorously) state that

converges because

does?

Thanks.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Really, really quick

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

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Really, really quick

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

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.

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.

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"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..."

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Really, really quick

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

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Really, really quick

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.