If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is

Sum \(= 1/2^1 + 1/2^2 + 1/2^3+1/2^4+... = 1\)
Sum = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = 1 ..............(I)

Note that the first term is 1/2 so the sum of the rest of the terms must be 1/2 to add up to a total of 1.
So 1/4 + 1/8 + 1/16 + 1/32 + ... = 1/2

Similarly, now the first term is 1/4 so the sum of the rest of the terms must be 1/4 too to get a sum of 1/2.

and so on...

Required sum \(= \frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+....\)

Required sum \(=(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) + (\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) + (\frac{1}{2^3}+\frac{1}{2^4}+...) + ...\)

Required sum \(= (1) + (\frac{1}{2}) + (\frac{1}{4}) + (\frac{1}{8})+ ...\)

From equation (I) above, we know that 1/2 + 1/4 + 1/8 + ... = 1
So Required sum = 1 + 1 = 2­

Another quirky series question is discussed here: 
https://youtu.be/KX8WNiyNUIo­