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

[meaw]

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 !

 

[HallsofIvy]

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.

 

[meaw]

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 :)

 

[genneth]

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.

 

[meaw]

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 !

 

[ssd]

The series is not absolutely convergent. Therefore rearrangement if its terms is not permissible.

 

[Gib Z]

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.

 

[meaw]

>> 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

 

[Gib Z]

Yup you are correct. My main point still holds though =]

 

[grmnsplx]

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

 

[LukeD]

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.

 

[Dchmm]

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.