August 29th, 2015, 06:08 PM  #21  
Senior Member Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus  Quote:
 
August 30th, 2015, 04:58 AM  #22 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 
That's an even simpler example, manishsqrt. But that sum DOES equal 3, doesn't it?
Last edited by Timios; August 30th, 2015 at 05:05 AM. 
August 31st, 2015, 07:30 PM  #23 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra  
September 2nd, 2015, 03:11 PM  #24 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 
I've been trying to discover the algebraic error in the analagous "proof" that mrtwhs provided. Is the error constituted in the supposition that one can subtract an infinite additive series with terms increasing numerically, from an identical infinite series and obtain zero as a result? Or is the error in supposing that there exists an infinite sum of terms with increasing numerosity that CAN be so subtracted? Or is the error ... ? (Please excuse my inadequacy in failing to express the above with better mathematical terminology.) 
September 2nd, 2015, 06:13 PM  #25 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
Your error is in supposing, without proof, that all infinite sums can be manipulated in the same way as finite sums. They can't. In particular, what you are doing is changing the order of the terms, which is only a valid operation for absolutely convergent series.
Last edited by v8archie; September 2nd, 2015 at 06:15 PM. 
September 2nd, 2015, 08:55 PM  #26 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
One very quirky example is the alternating harmonic series $\dfrac{1}{2} \dfrac{1}{3} + \dfrac{1}{4}  \dfrac{1}{5} + ...$ In it's usual form, it converges to $\ln 2$ but if the terms are rearranged, it can converge to ANY REAL NUMBER WE WANT IT TO! For example, say I want it to converge to $8$. I would start adding the positive terms of the sequence ($\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + ...$) until the sum goes over $8$. Then I start adding the negative terms until the sum goes under $8$. Then I start adding positive terms again until the sum goes over $8$. Repeat. This works because the series is not absolutely convergent. It is only conditionally convergent. If you rearrange a conditionally convergent series, you get a completely different result. 
September 3rd, 2015, 04:19 AM  #27 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
You can make any conditionally convergent series take any value you wish by rearranging the terms. The alternating harmonic series is not special in this regard.

September 3rd, 2015, 09:20 AM  #28 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timeywimey stuff.  

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