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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.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.

#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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