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#1 2013-02-09 22:49:33

Johnathon bresly
Guest

Convergence

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

#2 2013-02-10 03:43:18

bobbym
Administrator

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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.

#3 2013-02-12 00:39:33

scientia
Full Member

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Re: Convergence

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

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

Examples.

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

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

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