Σ2<sup>-n</sup> = Σn·2<sup>-n</sup>

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Σ2^-n = Σn·2^-n

Admitedly, this is a strange identity to contemplate: every term at the right is greater than the corresponding term on the left. How then the identity is possible?

The identity becomes more plausible with the remark that the summation on both sides starts with n = 0:

(1)	Σ_{n = 0} 2^-n = Σ_{n = 0} n·2^-n

which effectively kills the first term on the right. This becomes transparent if we write the series explicitly:

(1')	1 + 1/2 + 1/4 + 1/8 + ... = 0·1 + 1·1/2 + 2·1/4 + 3·1/8 + ...

Is it now less surprising?

We shall assume that the two series converge. (This is shown by the means taught in the beginning Calculus course. We skip this here. The point is however similar to that made in the discussion of the identity 0.999 ... = 1. In the presence of convergence simple algebraic manipulations of a series lead to valid results that may not be true for the divergent series.)

Thus denote

S = 1 + 1/2 + 1/4 + 1/8 + ... and
T = 1/2 + 2/4 + 3/8 + ...

It's actually more convenient to work with the summation symbol Σ:

T	= 1/2 + 2/4 + 3/8 + ...
	= Σ_{n = 0}n·2^-n
	= Σ_{n = 1}n·2^-n
	= Σ_{n = 1}(n - 1)·2^-n + Σ_{n = 1}2^-n
	= ½Σ_{n = 1}(n - 1)·2^-(n-1) + ½Σ_{n = 1}2^-(n-1)
	= ½Σ_{n = 0}n·2^-n + ½Σ_{n = 0}2^-n
	= ½S + ½T.

Which implies

S = S/2 + T/2, or S/2 = T/2

proving the promised S = T. As S is the sum of a geometric series with the factor q = 1/2,

S	= 1 / (1 - q)
	= 1 / (1 - 1/2)
	= 1 / 1/2
	= 2.

Therefore, 2 = S = T.

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