1+4+16+64+256+... = - 1/3

1 = 1
1+4 = 5
1+4+16 = 21
1+4+16+64 = 85

...and so forth.  Each time we add a term, it's the previous sum plus an even larger number.  With 13 terms, it's already up to about 22 million.  And yet, if we add enough of these series together, we should get -1/3.

Why?

Taylor series expansion for 1/(1-x^2) is

SUM (2^(2n)) for n=0 to n = infinity.  The earlier we stop (lower n) the sum.  So plug in 2.  The numbers in the series sum are 1, 4, 16, 64, etc.  It's always going to get bigger.  But what about plugging in the value directly. 

2^2 = 4

1-4 = -3

1/ (-3) = -1/3

Hence, infinity = -1/3?  But if you try it with another number, you get another series.  What's going on here?

Taylor series expansion for 1/(1-x^2) is SUM (2^(2n)) for n=0 to n = infinity.
Proof?

1. (2^0)²
2. + (2^1)²
3. + (2^2)²
4. + ...
5. = profit

>>2
Use a telescoping sum trick to show that 1/(1-r) is the sum of (r^n) for n from 0 to infinity when |r| < 1.  >>1 used the formula with r=4.  Fail.

>>4
Oh, right. I was confused because I didn't see the x reappear in his sum, but now I see he already replaced it with the 2 which he mentioned in the next step. Yeah, >>1 is a pretty massive fail.

>>4
That's the whole question.  Why does it fail?  In particular, what is it about the singularity that makes it fail?

because 4 is not less than 1

>>6
Perhaps a brief derivation will help:
1-x^N = (1-x) + (x-x^2) + (x^2-x^3) + ... + (x^(N-1)-x^N))
 = (1-x)(1+x+x^2+ ... +x^(N-1))
-> 1+x+x^2+...+x^(N-1) = (1-x^N)/(1-x).
Now, let N -> +Inf and consider what happens.  If |x| < 1, then x^N -> 0, so you simplify the RHS to get nice formula.  If |x| > 1, then x^N diverges, you can't evaluate either side of the equation, so you get no formula.  HTH HAND.