Investigating a Puzzling Equation: ln(2) = 1?

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Discussion Overview

The discussion revolves around the equation ln(2) = 1, specifically examining the validity of a power series expansion of ln(1+x) and the implications of manipulating divergent series. Participants explore the convergence of series, the rearrangement of terms, and the nuances of Taylor expansions.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims ln(2) can be expressed as ln(1+1) and derives a series that leads to the conclusion 2 = 1.
  • Another participant points out that the series for ln(1+x) at x=1 does not converge, challenging the validity of the initial calculation.
  • A different participant acknowledges the divergence of part of the series but argues that the overall expression remains convergent.
  • One participant emphasizes the importance of the ordering of terms in Taylor expansions, noting that rearranging divergent sums can lead to misleading results.
  • Another participant offers an alternative approach to derive ln(2) using a different manipulation of series, suggesting it leads to a correct conclusion.
  • Concerns are raised about the lack of absolute convergence in the series, which affects the permissibility of rearranging terms.
  • Participants discuss the interval of convergence for the series, with some asserting it is (0,1] while others suggest it is (-1,1].
  • One participant humorously illustrates the pitfalls of manipulating infinite series by presenting a paradoxical result.
  • Another participant asserts that the series 1 - 1 + 1 - 1... converges to 1/2, but emphasizes that the divergence of the harmonic series prevents valid rearrangements.
  • Concerns are raised about the validity of constructing a series from a divergent series, noting that it cannot be completed properly.

Areas of Agreement / Disagreement

Participants express disagreement on the convergence and manipulation of series, with multiple competing views on the validity of the calculations and the implications of divergent series. No consensus is reached regarding the conclusions drawn from the series manipulations.

Contextual Notes

Limitations include the dependence on the convergence properties of the series discussed, the potential for misinterpretation of divergent sums, and the unresolved nature of the mathematical steps involved in the manipulations.

meaw
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Tell me what is wrong with this :)

ln (2) = ln( 1 +1 ) and the power series expansion of ln(1+x) for x=1 gives

ln(2) = 1 - 1/2 + 1/3 -1/4 + 1/5 + ...
= (1 + 1/2 + 1/3 + 1/4 + 1/5 + ... )
- 2 . ( 1/2 + 1/4 + 1/6 + .....)

= (1 + 1/2 + 1/3 + 1/4 + 1/5 + ... )
- (1 + 1/2 + 1/3 + ...)
= 0 = ln (1)

=> 2 = 1 ! huh !
 
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The interval of convergence for ln(1+ x) at x= 1 is (0, 1). The series you get at x= 1, 1+ 1/2+ 1/4+ ... does not converge so your calculation is invalid.
 
you are partially right. you are right n the sense that (1 + 1/2 + 1/3 + 1/4 ...) is divergent. but you are worng in the sense that you only considered that portion of the sum, the whole thing together is not divergent. (x + x) - 2x : is convergent even if (x+x) is divergent :)
 
The definition of a Taylor expansion relies on the ordering of the terms. When manipulating infinite sums, it is no longer true that addition commutes. This is a well-known issue, and many brilliant minds have worked to understand it. A quick Google for "divergent sums" will get you a long way towards what people have already thought of.
 
agreed. a more elegant answer is here :)

ln(2) = 1 - 1/2 + 1/3 -1/4 + 1/5 - 1/6 ... +- 1/2n where n -> inf
= (1 + 1/2 + 1/3 + 1/4 + 1/5 + + ... 1/2n where n -> inf
- 2 . ( 1/2 + 1/4 + 1/6 + ... this has only n terms )

= (1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 1/2n)
- (1 + 1/2 + 1/3 + ... + 1/n)

= 1/n+1 + 1/n+2 + 1/n+3 ... 1/2n

= intergral of (1/1+x) from 0 to 1 , from Newton's summation formula of definite integral

= ln (1+x) from 0 to 1

= ln 2

lol !
 
The series is not absolutely convergent. Therefore rearrangement if its terms is not permissible.
 
The interval of convergence for ln (1+x)'s series is (0,1]. When x=1, the series is famously the "alternative harmonic series" and equal to ln 2.
 
>> The interval of convergence for ln (1+x)'s series is (0,1].

I thought the interval is (-1,1], open on -1, closed on +1
 
Yup you are correct. My main point still holds though =]
 
  • #10
yeah be careful with the infinite series.
0=+1 + (-1) right?
so
0= 1-1+1-1+1-1+1-1...
= 1+(-1+1)+(-1+1)+(-1+1)...
= 1 +0+0+0+0...
=1

which is a load of crap
 
  • #11
really now, everyone knows that 1 - 1 + 1 - 1 ... = 1/2

Of course, when I write it like that, I mean ((((1 - 1) + 1) - 1) + 1) and so on

of course (1-1) + (1-1) + (1-1) ... = 0, and nothing else. What else could it consistently be?

Anyway, whether or not 1 - 1/2 + 1/3 - 1/4 ... converges doesn't actually matter here. The fact is that 1 + 1/2 + 1/3 + 1/4 + ... does not converge, so the first series I wrote down can't be rearranged or you can do so to get any number that you want at all.
 
  • #12
The series you construct from the first series is only half as long. Therefore you can't do it. Half the subtraction sum can't be finished.
 

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