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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: 87,164

Re: Convergence


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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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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