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#1 2013-02-08 23:49:33

Johnathon bresly
Guest

Convergence

What is the difference between absolute and conditional convergence?[examples will be appreciated]

#2 2013-02-09 04:43:18

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,777

Re: Convergence


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#3 2013-02-11 01:39:33

scientia
Member
Registered: 2009-11-13
Posts: 222

Re: Convergence

The sequence gif.latex?\sum_na_n is absolutely convergent iff both gif.latex?\sum_{n=0}^{\infty}a_n and gif.latex?\sum_{n=0}^\infty|a_n| converge.

It is conditionally convergent iff gif.latex?\sum_{n=0}^{\infty}a_n converges while gif.latex?\sum_{n=0}^\infty|a_n| diverges.

Examples.

gif.latex?\sum_n\frac{(-1)^n}{2^n} is absolutely convergent. We have gif.latex?\sum_{n=0}^\infty\frac{(-1)^n}{2^n}=1-\frac12+\frac14-\frac18+\cdots=\frac23 and gif.latex?\sum_{n=0}^\infty\left|\frac{(-1)^n}{2^n}\right|=1+\frac12+\frac14+\cdots=2.

gif.latex?\sum_n\frac{(-1)^n}n is conditionally convergent. We have gif.latex?\sum_{n=0}^\infty\frac{(-1)^n}n=1-\frac12+\frac13-\frac14+\cdots=\ln2 while gif.latex?\sum_{n=0}^\infty\left|\frac{(-1)^n}n\right|=1+\frac12+\frac13+\cdots is divergent.

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