Adding 2^n up to infinity

S = 1+2+4+8+16+32+64.................. up to infinity S = 0+1+2+4+8+16+32+...................up to infinity

subtracting both S-S = 0 = 1+1+2+4+8+16+32....................... = 1+ S therefore, S = -1

How come is it possible?

#Math

Note by Rahul Dandwate
8 years, 1 month ago

No vote yet
2 votes

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Comments

To understand what is happening, let's modify the problem a little bit. Define S(n)=1+2+⋯+2n S(n) = 1 + 2 + \cdots + 2^n for all nonnegative integers n n . In the original post, you are calculating "S - S" by summing the differences of the kthk^{\rm th} and (k+1)th(k+1)^{\rm th} terms of S, for each k=1,2,3,…k = 1, 2, 3, \ldots . If you were to do this with S(n) S(n) , you would see that we get S(n)−S(n)=(1−0)+(2−1)+(4−2)+⋯+(2n−2n−1)+(0−2n)=1+1+2+⋯+2n−1−2n. \begin{aligned} S(n) - S(n) &= (1 - 0) + (2 - 1) + (4 - 2) + \cdots + (2^n - 2^{n-1}) + (0 - 2^n) \\ &= 1 + 1 + 2 + \cdots + 2^{n-1} - 2^n. \end{aligned} As you can see, the final term is negative, and this results in the identity 2n−1=1+2+4+⋯+2n−1=S(n). 2^n - 1 = 1 + 2 + 4 + \cdots + 2^{n-1} = S(n). There is no contradiction.

Now, as n n tends to infinity, we see that S(n)=2n−1 S(n) = 2^n - 1 also tends to infinity, not −1 -1 . The flaw in the above "proof" is that writing S−S S - S as an infinite series "hides" the existence of the term 0−2n 0 - 2^n in the finite series, which tends to −∞ - \infty as n→∞ n \to \infty . It is because this term does not vanish to zero that the difference S - S cannot be said to be equal to 1 + S, and that is where your calculation is incorrect.

hero p. - 8 years, 1 month ago

i think you can't substract both S if both of them are infinity... after all, 1+1+2+4+8+16+32+... can't be 0....

リチャット ウェリアント - 8 years, 1 month ago
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