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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, 9 months ago

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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.

- 4 years, 9 months 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....

- 4 years, 9 months ago