Proof that an Infinite Series Adds to 1 We will prove that: $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}... = 1$ (exactly) Proof: (1) Let n = $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ Multiply each side of (1) by 2 (2) 2n = $\displaystyle 1 + \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ Subtract each side of line (1) from each side of line (2) (3) n = 1 One can also use this technique to prove that .999... is exactly equal to 1. 
Be careful when using proofs like this. Saying "Let $n = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ...$" implies that the series converges, which may not be the case. There are (false) proofs that the series $1  1 + 1  1 + ...$ converges to $0$, $1$ and $\dfrac{1}{2}$ using this principle, but the series does not converge at all. If you want to formalise this proof, you have to first prove that the series converges. 
This is, in fact, a "geometric series". A geometric series is any series of the form $\displaystyle \sum_{n=0}^\infty ar^n= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot$. One can prove that a geometric series converges if and only if r< 1 and then converges to $\displaystyle \frac{a}{1 r^2}$. Here, we can write the series as $\displaystyle \frac{1}{2}\left(1+ \frac{1}{2}+ \frac{1/4}+ \cdot\cdot\cdot\right)$ so 1/2 times a geometric sequence with a= 1 and r= 1/2. Since 1/2< 1, this converges to $\displaystyle \frac{1}{2}\frac{1}{1 \frac{1}{2}}= \frac{1}{2}\frac{2}{2 1}= 1$. (This is also a geometric series with a= 1/2 and r= 1/2 so converges to $\displaystyle \frac{\frac{1}{2}}{1 \frac{1}{2}}= \frac{\frac{1}{2}}{\frac{1}{2}}= 1$) (It can also be thought of as a geometric series with a= 1 and r= 1/2 with the first "a" term missing. The geometric series with a= 1 and r= 1/2, $\displaystyle 1+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \cdot\cdot\cdot$ converges to $\displaystyle \frac{1}{1 \frac{1}{2}}= 2$ and subtracting the missing first term, "1" gives a sum of 1.) 
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How in the world did that "2" sneak in there?! Yes, of course. Thank you. 
Perhaps I'm stupid, but I see no necessity of first proving that the series converges. Whether it converges or not, the proof I provided is purely algebraic: Multiplying each side of an equation by 2, and Subtracting the same thing from each side of an equation. 
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For example: S = 1 + 2 + 3 + 4 + ... 2S = 2 + 4 + 6 + 8 + .... S = 2S  S = 1  3  5  7  ... But can we say that 1 + 2 + 3 + 4 + ... = 1  3  5  7  .... ? Dan 
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Suppose that $$s = 1  1 + 1  1 + \cdots$$Then $$s = 1 + 1  1 + 1 \cdots = 1 + s$$Thus $$2s = 1 \implies \boxed{s = \tfrac12}$$ This uses all the same techniques that yours does, but is patently not true. The partial sums never get closer to $\frac12$ at all! What you are really saying is that "I get the right answer, so why should I prove that the method is sound". But that means that what you have is not a proof that the sum is what you say it is because applied to another sequence we don't know whether the result is correct or not. We have to check by another method. 
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S = 1 + (2  2) + (3  4) + (4  6) + ... S = 1 + 0  1  2  3  ... S = 1  S S = 1/2 
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2S= 2 + 4 + 6 + 8 + ... S= 1 + 2 + 3 + 4 + ... Subtract both sides and you have: S = 1 + 2 + 3 + 4 + ... 
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