4 and . . . = 0(simonsingh.net)

13 pointsby ColinWright11 anni fa13 comments

Paradox: 1 – 1/2 and 1/3 – 1/4 and . . . = 0

http://www.simonsingh.net/Simpsons_Mathematics/1-12-13-14-0/

5 comments

It's interesting to note that since it is conditionally convergent, one can in fact rearrange the order of the terms in the series 1 - 1/2 + 1/3 - 1/4... to obtain a series that does in fact converge to 0.

http://en.wikipedia.org/wiki/Riemann_series_theorem

Patient011 anni fa

Yeah that's what I thought the article was going to be about. By re-arranging terms you can make it convert to any real number.

gglon11 anni fa

And 1 + 2 + 3 + 4 + ⋯ = ...

bengali311 anni fa

where is the easiest place on the web to graph this?

ColinWright11 anni fa

What do you want to graph?

The first expression, A1, is:

    sum_{x=1}^{oo} 1/x

If you convert that into a function:

    S(n) = sum_{x=1}^n 1/x

then you'll find that S(n) ~ ln(n)-c where c is a constant.

You can graph these sorts of things at Wolfram Alpha, or even just type the expression into Google and often you'll get something useful. Here's an example:

https://news.ycombinator.com/item?id=9393392

bengali311 anni fa

Thanks, in my messing around with WA, I wasnt getting a chart using the limit function, but now i realized I wanted the summation. I havent done much graphing online.

TBH I was just curious how the graphs looked taking the positives and negatives independently.

sum_{x=1}^{oo} 1/x

vs

sum_{x=1}^{oo} (-1)1/x

vs

sum_{x=1}^{oo} (-1)1/(x+1)

ColinWright11 anni fa

I suspect you have problems there with asterisks for multiplication. Edit it to use dot instead of star, then I might be able to reply sensibly.

lorddoig11 anni fa

8pts, 6 comments and already the site's buckled.

Sometimes this place makes me think that not putting a caching proxy in front of your blog should be a crime.

slevin06311 anni fa

A1 and A2 have different limits. It will be clear if the upper limit is changed to something other than infinity. Lets say A1 is sum of 1/x, x=1 to 10. For A1 and A2 to be equal, A2 will have limits x=1 to 5. Thus, you cannot strike of sum of (1/x) on both sides in the equation, A1=D.

ColinWright11 anni fa

  > A1 and A2 have different limits. It will be clear
  > if the upper limit is changed to something other
  > than infinity.

While it might be true that A1 and A2 have different limits, the reason you give is not sufficient to prove it. Consider the following:

  1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

and

  1/2 + 1/8 + 1 + 1/32 + 1/128 + 1/4 + ...

In this second case I'm taking two odd powers then an even power, and so on.

If your argument were valid then these would have different limits, because if you cut them both off after some number of terms the totals will always be different. However, these two series in fact have the same limit.

And in fact A1 and A2 have the same limit, so that's not where the problem lies.

slevin06311 anni fa

>For A1 and A2 to be equal, A2 will have limits x=1 to 5.

I should have added 'For A1 and A2 to be equal' at the start. If you want them to be equal, they need to have different limits and if they have same limits, they are not equal.

ColinWright11 anni fa

I'm finding that completely impossible to parse.

In particular:

    If you want them to be equal, they need
    to have different limits and if they have
    same limits, they are not equal.

That seems to be completely wrong. Consider these:

    sum_{x=0}^oo (-1)^x (1/x)

and

    sum_{y=0}^oo [ 1/(2y) - 1/(2y+1) ]

These are equal, and yet replacing the "oo" with n always gives partial sums that are different. So I guess I just don't understand what you are trying to say. Perhaps you could be more complete and explicit.

[deleted]11 anni fa

dhanrajm11 anni fa

Exactly. If you use different variable at two sides..say 'x' and 'n' , it will be more clear.