5 ms·

1 - 2 + 3 - 4 + 5 - 6 + ... made easy

49 points by vinutheraj 17y ago

scythe 17y ago
I personally found the part of the article about Abel summation[1] much more enlightening. It really seems to indicate why 1 - 2 + 3 ... = 1/4 instead of just showing that 1/4 arises if you look at the sums a certain way. [1]: http://en.wikipedia.org/wiki/1_−_2_%2B_3_−_4_%2B_·_·_·#Abel_summation http://en.wikipedia.org/wiki/1_−_2_%2B_3_−_4_%2B_·_·_·#Abel_...
tsally 17y ago
I alway prefer this type of approach for solving problems. Even when formal proofs are required, I usually get the answer in this way and then reverse engineer the proof. Another example (the sum of 1/2 + 1/4 + 1/8 + 1/16...): http://media-2.web.britannica.com/eb-media/79/26979-004-CF3F4DA2.gif http://media-2.web.britannica.com/eb-media/79/26979-004-CF3F...
- eru 17y ago
  Almost all mathematical proofs are reverse engineered.
spicyj 17y ago
I find it more helpful to note that 1/(1 + x)^2 = 1 - 2x + 3x^2 - 4x^3 + ... for |x| < 1 and then to extend the summation to other values of x.
jaydub 17y ago
http://en.wikipedia.org/wiki/Telescoping_series http://en.wikipedia.org/wiki/Telescoping_series
antimora 17y ago
Wolfram Alpha has made even simpler, and (more correct proof) http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+.. http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-....
- tectonic 17y ago
  Wolfram Alpha continues to impress me.
  - hyperbovine 17y ago
    But not me http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+7.. http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-.... .
    - vinutheraj 17y ago
      Adding a '+' to the end, before the '..' makes it work again, maybe it needs an operand and the '..' to show that it is an infinite series, maybe ! http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+7%2B. http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-....
- antimora 17y ago
  hmmm ... for whatever reasons the link was broken. Trying it again: "http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+... http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-...
  - antimora 17y ago
    here is a PDF snapshot. "http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP55619821efaa27908he00002if937a6faf7b84c&s=31&i=1+-+2+%2B+3+-+4+%2B+5+-+6+%2B+... http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP55619821e...
- zck 17y ago
  Although I'm not a fan of using i in a mathematical formula like this. When I first saw it, it looked like they meant sqrt(-1).
thras 17y ago
It diverges under the ordinary definitions of infinite series, but one can generalize the notion of summation a little bit to allow for certain divergent series to have finite sums.
mitko 17y ago
All such attempts for summation are ill-defined. The Abel summation is actually not a property - it is a functional. That said you take the function f(n)=<n-th summand>. Then you can define AS(f), but it shares very few properties with normal summation. If you are not careful you can easily reach contradiction. For example 1/4 = 1 -2 +3 -4 +5 -6 +7 ... = //comment: 1-2-4+5=0, 3-6+9-12... = 3\sum i(-1)^i= 3* 1/4 Contradiction! Edit: The original contradiction example was more complicated.
- BearOfNH 17y ago
  It's a neat example, but surely you know you can't rearrange terms of an infinite sum unless it is absolutely convergent. That's not the case here.
edw519 17y ago
I don't know if a real mathematician would approve of my method, but at least it's easy to understand. The good mathematician may not approve. The better mathematician probably would approve. The best mathematician would seek and codify a principal to describe the general case of the instance you cite.
- eru 17y ago
  If she has nothing better to do at the moment.

49 points by vinutheraj 17y ago

I personally found the part of the article about Abel summation[1] much more enlightening. It really seems to indicate why 1 - 2 + 3 ... = 1/4 instead of just showing that 1/4 arises if you look at the sums a certain way. [1]: http://en.wikipedia.org/wiki/1_−_2_%2B_3_−_4_%2B_·_·_·#Abel_summation http://en.wikipedia.org/wiki/1_−_2_%2B_3_−_4_%2B_·_·_·#Abel_...

tsally 17y ago

I alway prefer this type of approach for solving problems. Even when formal proofs are required, I usually get the answer in this way and then reverse engineer the proof. Another example (the sum of 1/2 + 1/4 + 1/8 + 1/16...): http://media-2.web.britannica.com/eb-media/79/26979-004-CF3F4DA2.gif http://media-2.web.britannica.com/eb-media/79/26979-004-CF3F...

eru 17y ago
Almost all mathematical proofs are reverse engineered.

spicyj 17y ago

I find it more helpful to note that 1/(1 + x)^2 = 1 - 2x + 3x^2 - 4x^3 + ... for |x| < 1 and then to extend the summation to other values of x.

jaydub 17y ago

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

antimora 17y ago

Wolfram Alpha has made even simpler, and (more correct proof) http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+.. http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-....

tectonic 17y ago
Wolfram Alpha continues to impress me.
- hyperbovine 17y ago
  But not me http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+7.. http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-.... .
  - vinutheraj 17y ago
    Adding a '+' to the end, before the '..' makes it work again, maybe it needs an operand and the '..' to show that it is an infinite series, maybe ! http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+7%2B. http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-....
antimora 17y ago
hmmm ... for whatever reasons the link was broken. Trying it again: "http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-6+%2B+... http://www.wolframalpha.com/input/?i=1+-+2+%2B+3+-+4+%2B+5+-...
- antimora 17y ago
  here is a PDF snapshot. "http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP55619821efaa27908he00002if937a6faf7b84c&s=31&i=1+-+2+%2B+3+-+4+%2B+5+-+6+%2B+... http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP55619821e...
zck 17y ago
Although I'm not a fan of using i in a mathematical formula like this. When I first saw it, it looked like they meant sqrt(-1).

thras 17y ago

It diverges under the ordinary definitions of infinite series, but one can generalize the notion of summation a little bit to allow for certain divergent series to have finite sums.

mitko 17y ago

All such attempts for summation are ill-defined. The Abel summation is actually not a property - it is a functional. That said you take the function f(n)=<n-th summand>. Then you can define AS(f), but it shares very few properties with normal summation. If you are not careful you can easily reach contradiction. For example 1/4 = 1 -2 +3 -4 +5 -6 +7 ... = //comment: 1-2-4+5=0, 3-6+9-12... = 3\sum i(-1)^i= 3* 1/4 Contradiction! Edit: The original contradiction example was more complicated.

BearOfNH 17y ago
It's a neat example, but surely you know you can't rearrange terms of an infinite sum unless it is absolutely convergent. That's not the case here.

edw519 17y ago

I don't know if a real mathematician would approve of my method, but at least it's easy to understand. The good mathematician may not approve. The better mathematician probably would approve. The best mathematician would seek and codify a principal to describe the general case of the instance you cite.

eru 17y ago
If she has nothing better to do at the moment.