

Paradox: 1 – 1/2 and 1/3 – 1/4 and . . . = 0 - ColinWright
http://www.simonsingh.net/Simpsons_Mathematics/1-12-13-14-0/

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gglon
And 1 + 2 + 3 + 4 + ⋯ = ...

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lorddoig
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.

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andbberger
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](http://en.wikipedia.org/wiki/Riemann_series_theorem)

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Patient0
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.

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bengali3
where is the easiest place on the web to graph this?

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ColinWright
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](https://news.ycombinator.com/item?id=9393392)

~~~
bengali3
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)

~~~
ColinWright
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.

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slevin063
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.

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ColinWright

      > 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.

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slevin063
>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.

~~~
ColinWright
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.

