My Math Forum Proof that an Infinite Series Adds to 1

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

August 29th, 2015, 06:08 PM   #21
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 Originally Posted by Timios We will prove that: $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}... = 1$ (exactly) Proof: (1) Let n = $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ Multiply each side of (1) by 2 (2) 2n = $\displaystyle 1 + \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ Subtract each side of line (1) from each side of line (2) (3) n = 1 One can also use this technique to prove that .999... is exactly equal to 1.
it doesn't matter, this could result into 3 if you place 3 instead of 1 in numerator of each term on right hand side. Its just a trick no theorem.

 August 30th, 2015, 04:58 AM #22 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 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
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 Originally Posted by mrtwhs Put yourself in for a Fields Medal. You've shown that a bunch of positive numbers adds to $\displaystyle -3$.
$\zeta(-1) = -{1 \over 12}$

 September 2nd, 2015, 03:11 PM #24 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 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: 7,671 Thanks: 2651 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: 7,671 Thanks: 2651 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
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Quote:
 Originally Posted by Azzajazz 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 $\ln 2$
Just a small correction. $\displaystyle ln(2) = 1 - \frac{1}{2} + \frac{1}{3} + ...$.

-Dan

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