# Infinity puzzle

## Main Question or Discussion Point

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
Homework Helper
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 !

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

Gib Z
Homework Helper
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

Gib Z
Homework Helper
Yup you are correct. My main point still holds though =]

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

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.

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.