.Unraveling the Mystery of 1=2: A Logarithmic Proof

[fourier jr]

everybody seems to know at least one of the "proofs" that 1=2 or 1=-1, etc but i had never seen this one before. check it out:
everybody knows that
$$log(1+x) = x-\frac{x^2}{2} + \frac{x^3}{3} - ...$$

plug in x=1 & the series converges & we get

log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - ...
2log2 = 2 - 1 + 2/3 - 1/2 + 2/5 - 1/3 + 2/7 - ...

take the terms together which have a common denominator (ie simplify) & we get

2log2 = 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 - ... = 1 - 1/2 + 1/3 -1/4 + ... = log2

hence 1 = 2
QED



here's a similar one
log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...
= (1 + 1/3 + 1/5 + 1/7 + ...) - (1/2 + 1/4 + 1/6 + 1/8 + ...)
= {(1 + 1/3 + 1/5 + ...) + (1/2 + 1/4 + 1/6 + ...)} - 2(1/2 + 1/4 + 1/6 + ...)
= (1 + 1/2 + 1/3 + ...) - (1 + 1/2 + 1/3 + ...)
= 0

i guess the problem must have something to do with the 'simplification' & doing something to an infinite sum. off the top of my head those are my guesses

 

[shmoe]

In the first one, you are rearranging the terms, does that make you at all queasy?

In the second one, you're splitting the limit into the sum of two limits from the first line to the second, when is this allowed?

 

[dextercioby]

At the end you're subtracting $\infty -\infty$.

Daniel.

 

[Zurtex]

You do know about the Riemann Series Theorem right?

http://mathworld.wolfram.com/RiemannSeriesTheorem.html

 

[saltydog]

How you arrange the terms of an alternating series matters. In fact, you can re-arrange them to make the series converge to any number you want.

Edit: Jesus Zurtex. Didn't realize the Reimann Series Theorems was about that.

 

[arildno]

Still, fourierjr has certainly provided one of the more subtler "proofs" of 2=1.

I think most who haven't studied series would be fooled by this argument.

 

[fourier jr]

i first learned about the "riemann series theorem" about a year & a half ago but didn't know it had a name. schmoe's msg was enough for me to see the problem. I'm surprised i didn't see it before. i even typed it up & still didn't see it.

here's another one, by d'alembert.
everyone knows that if the product of 2 numbers equals the product of 2 other numbers, the numbers will bee in proportion. from the definition of proportion if the 1st number is greater than the 2nd then the 3rd will be greater than the 4th, ie if ad=bc, then a:b = c:d and if a>b then c>d. so far so good. now set a=d=1 and b=c=-1 and there are 4 number that satisfy the relation ad=bc and a>b & by the proposition c>d, ie -1>1. QED

 

[krab]

if ad=bc, then a:b = c:d and if a>b then c>d. so far so good.

Not good at all, since not true. E.g. a=2,b=1,c=-4,d=-2 violates your statement. If all a,b,c,d are positive, then what you say is true.

 

[fourier jr]

lol note to self: must work on critical-thinking skills...

 

[Icebreaker]

I don't understand the Riemann Series Theorem; if a series can be rearranged into converging into any value or diverging, then nothing can be said of the series. That doesn't make sense.

 

[Hurkyl]

The ordering is part of the series. You rearrange it, you get a different series.

 

[fourier jr]

I don't understand the Riemann Series Theorem; if a series can be rearranged into converging into any value or diverging, then nothing can be said of the series. That doesn't make sense.

yes it does. if you've got a series that converges conditionally, you can make it sum to anything you want. say you want it to sum to 5. add up just enough positive terms so that your partial sum is a bit more than 5, then start adding negative terms until you go less than 5... & so on. keep going back & forth like that & eventually you'll be inside a neighbourhood around 5 forever (after the sum of some N terms). so you can make a conditionally-convergent series converge to anything by arranging the terms the right way.

another fallacy, this one from gt watson:
consider the identity $$\sqrt{x-y} = i\sqrt{y-x}$$
fix x=a & y=b & get $$\sqrt{a-b} = i\sqrt{b-a}$$
now fix x=b & y=a & get $$\sqrt{b-a} = i\sqrt{a-b}$$

multiply them together & get that
$$\sqrt{b-a}\sqrt{a-b} = i^2\sqrt{b-a}\sqrt{a-b}$$
ie. 1 = -1 QED

although an identity is the same no matter what numbers you put in, the problem with this one is that x & y are FIXED at x=a & y=b. so they can't be changed to something else. haha i think i got that one

 

[Icebreaker]

The ordering is part of the series. You rearrange it, you get a different series.

That's new to me. Time to hit the textbooks.

 

[fourier jr]

That's new to me. Time to hit the textbooks.

i know it's listed as a problem in pfafenberger/johnsonbaugh's foundations of mathematical analysis, in the section on conditional convergence. it's probably in baby rudin & some calculus books.

 

[shmoe]

i know it's listed as a problem in pfafenberger/johnsonbaugh's foundations of mathematical analysis, in the section on conditional convergence. it's probably in baby rudin & some calculus books.

Did you take the course with Pfaff? He first presented it as an impressive trick for parties. Had us give him our favorite constant, I think I said pi^2/6 or something, then he proceeded to show how the alternating harmonic series could be rearranged to get this. It impressed me, but your average bloke off the street-doubtful. I'm still not sure what kind of parties he goes to.

This should be in any intro analysis book and I'd optimistically hope at least mentioned in every calculus book, but I find that doubtful.

 

[fourier jr]

yeah i had pfaffenberger. i don't remember him mentioning anything about any parties though.