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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?

Note by Rahul Dandwate
4 years, 6 months ago

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2 votes

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

Now, as \( n \) tends to infinity, we see that \( S(n) = 2^n - 1 \) also tends to infinity, not \( -1 \). The flaw in the above "proof" is that writing \( S - S \) as an infinite series "hides" the existence of the term \( 0 - 2^n \) in the finite series, which tends to \( - \infty \) as \( n \to \infty \). 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. - 4 years, 6 months ago

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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 - 4 years, 6 months ago

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