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 July 11th, 2012, 11:29 AM #1 Newbie   Joined: Jun 2012 Posts: 26 Thanks: 0 Infinite Series If you have an infinite series, does it matter in what order you add the terms?
 July 11th, 2012, 11:32 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Infinite Series No, the commutative property of addition allows the addends to be arranged in any possible order.
July 11th, 2012, 12:27 PM   #3
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Re: Infinite Series

Quote:
 Originally Posted by Mathforum1000 If you have an infinite series, does it matter in what order you add the terms?
It might. If the series is absolutely convergent, then it does not matter. If it is conditionally convergent it will.

An example (conditionally convergent) 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ..... Rearranging terms can produce any answer.

July 11th, 2012, 04:54 PM   #4
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Re: Infinite Series

Hello, Mathforum1000!

Quote:
 If you have an infinite series, does it matter in what order you add the terms?

If the terms are all positive, the answer is Yes.
If it is an alternating series, the answer is Maybe.

There is a classic fallcy that depends on overlooking this fact.
I believe this is what mathman referred to.

$\text{We know that: }\:\ln(1+x) \:=\:x\,-\,\frac{x}{2}\,+\,\frac{x}{3}\,-\,\frac{x}{4}\,+\,\frac{x}{5}\,-\,\frac{x}{6}\,+\,\cdots$

$\text{Let }x\,=\,1.
\;\;\text{W\!e have: }\:\ln(2) \;=\;1\,-\,\frac{1}{2}\,+\,\frac{1}{3}\,-\,\frac{1}{4}\,+\,\frac{1}{5} \,-\,\frac{1}{6}\,+\,\cdots$

[color=beige]. . . . . . . . . . . . . .[/color]$=\;\underbrace{\left(1\,+\,\frac{1}{3}\,+\,\frac{1 }{5}\,+\,\cdots\,\right)}_P \,-\, \underbrace{\left(\frac{1}{2}\,+\,\frac{1}{4}\,+\, \frac{1}{6}\,+\,\cdots\,\right)}_N$

$\text{Add and subtract }N:$

$\ln(2) \;=\;\left(1\,+\,\frac{1}{3}\,+\,\frac{1}{5}\,+\,\ cdots\,\right)\,+\,\underbrace{\left(\frac{1}{2}\, +\,\frac{1}{4}\,+\,\frac{1}{6}\.+\,\cdots\,\right) }_{+\,N}\,-\,\left(\frac{1}{2}\,+\,\frac{1}{4}\,+\,\frac{1}{6 }\,+\,\cdots\,\right)\,-\,\underbrace{\left(\frac{1}{2}\,+\,\frac{1}{4}\,+ \,\frac{1}{6}\,+\,\cdots\,\right)}_{-\,N}$

[color=beige]. . . . [/color]$=\;\left(1\,+\,\frac{1}{2}\,+\,\frac{1}{3}\,+\,\fr ac{1}{4}\,+\,\frac{1}{5}\,+\,\frac{1}{6}\,+\,\cdot s\,\right)\,-\,2\,\!\left(\frac{1}{2}\,+\,\frac{1}{4}\,+\,\frac {1}{6}\,+\,\cdots\,\right)$

[color=beige]. . . . [/color]$=\;\left(1\,+\,\frac{1}{2}\,+\,\frac{1}{3}\,+\,\cd ots\,\right) \,-\,\left(1 \,+\,\frac{1}{2}\,+\,\frac{1}{3}\,+\,\cdots\,\righ t)$

[color=beige]. . . . [/color]$=\;\;0$

$\text{Therefore: }\:\ln\,\! 2 \:=\:0 \;\;\;\Rightarrow\;\;\;e^0 \,=\,2$

$\;\;\;\text{And once again: }\:1 \:=\:2$

July 12th, 2012, 04:25 AM   #5
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Re: Infinite Series

Quote:
 If you have an infinite series, does it matter in what order you add the terms?
$\text{Let the infinite series be } \sum_{n=a}^{b} f(n)$

$\text{The above statement is true iff }$

$\left | \sum_{n=a}^{b} f(n) \right | = \sum_{n=a}^{b} \left | f(n) \right |$

$\text{Otherwise, it is false because "-" is a non-commutative binary operation.}$

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