Maths with me guys...

[z3ro]

[Question]


What's the sum of     1 + 2 + 4 + 8 + 16 + ...  (continues to infinity with each number twice as large as the previous one)




Sum?    =  Infinity ∞ ?! Right?




....






But wait.... 


1 * X = X    for any X  [OK?]


so the sum is similar to (1)(1+2+4+8+16+...)  [OK?]


Now, One is the same as (two minus one)


1 = (2-1)  [OK?]


So.. the sum is still the same as (2-1)(1+2+4+8+16+...)


If we multiply.....     we get... ( 2+4+8+16+32+...... -1 + -2 + -4 + -8 + -16 + -32 +...)   We have


2 - 2 = 0
+4 + -4 =0
...etc...

Every number cancels out.. leaving... -1 






    (1+2+4+8+16+...) = -1

[p_2001]

no.... i'm not sure if this is a troll...

.. See, your cancellation has a flaw.
you are canceling the SECOND term of the negative series with first of the positive...

hence sum is
∞ - 1 = ∞ ...


you cannot forget the un-cancelled term.

[Snayler]

So.. the sum is still the same as (2-1)(1+2+4+8+16+...)


If we multiply.....     we get... ( 2+4+8+16+32+...... -1 + -2 + -4 + -8 + -16 + -32 +...)   We have


2 - 2 = 0
+4 + -4 =0
...etc...

Every number cancels out.. leaving... -1 

    (1+2+4+8+16+...) = -1

Your logic is flawed. If you multiply:
2[(2-1)(1+2+4+8+16+...)] =
2(2-1)*2(1+2+4+8+16+...) =
(4-2)*(2+4+8+16+32+...)
which is still infinity.

[Kulverstukas]

You put too many spaces out of excitement.

[DaNePaLI]

You put too many spaces out of excitement.

Lmfao. Exactly.

[karsa]

You can't just write one equation, take a part of it and rewrite it to suite your needs. Well... sometimes you can but this is not the case. Rules must be followed and (2-1)*(1+2+4+8+...) is still equal to ∞ no matter what.

[Jboeggs]

You are not multiplying, subtracting, or dividing. You are also not adding by a negative, which means the answer will always be a positive. The answer is always going to be infinite.