[zackattack]

You mean, = 1/(x-1) ;)

Can someone please post a non-visual proof of why this is the case? In the meantime, I am working on figuring out my own.

[jonsen]

You can get the proof idea from the picture:

  1/4 is 1/3 of 3/4
  1/4 of 1/4 is 1/3 of 3/4 of 1/4
  etc.

In math:

  1/4   = 1/3*3/4
  1/4^2 = 1/3*3/4*1/4
  1/4^3 = 1/3*3/4*1/4^2
  etc.

Summing equations:

  (1/4^1 + 1/4^2 + ...) = 1/3 * 3/4 * (1 + 1/4^1 + 1/4^2 + ...)
  <=>
  (1/4^1 + 1/4^2 + ...) = 1/3 * 3/4 + 1/3 * 3/4 * (1/4^1 + 1/4^2 + ...)
  <=>
  x = 1/3 * 3/4 + 1/3 * 3/4 * x
  <=>
  x - 1/4 x = 1/4
  <=>
  3/4 x = 1/4
  <=>
  x = 4/3 * 1/4
  <=>
  x = 1/3

[swapspace]

Infinite GP: a/(1-r)

[zackattack]

Ah, true.

http://en.wikipedia.org/wiki/Geometric_series#Formula

[I_got_fifty]

This wouldn't happen on Digg.

[ewjordan]

Not a rigorous proof (I'll leave that to further investigation):

S(n) = 1/n + 1/n^2 + ...

= 1/n ( 1 + 1/n + 1/n^2 + ...) <--needs more justification in a rigorous proof

S(n) = 1/n ( 1 + S(n) )

Simple algebra from here:

n * S(n) - S(n) = 1

S(n) = 1 / (n-1)

[jakkals]

Rats!

Yes, I did indeed mean 1/(x-1). Thanks for the correction.