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I remember that it equals 1, but I do not remember how to show the summation.

Chad
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  • It is a geometric series, so you can approach the limit as $k\to \infty$ using the standard formula. There are also many ways to derive the limit from "scratch". – hardmath Nov 28 '16 at 06:11
  • Possible duplicate of Infinite Geometric Series Formula Derivation – hardmath Nov 28 '16 at 06:13
  • @hardmath This is not off topic – Jacob Wakem Nov 28 '16 at 06:25
  • @Alephnull: My vote to close was based on the Question being a duplicate, consistent with my two Comments. That said, the category "off topic" was used here (and in many cases) because a majority of those who voted to close selected the "missing context" subcategory. The OP should have made the body of the Question a self-contained statement of the the problem for which help is wanted. – hardmath Nov 28 '16 at 12:13

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Start with the sum: $$\sum_{k = 2}^\infty\frac{1}{2^{n-1}} = \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots$$ Note that this is a geometric sum with common ratio $r = \frac{1}{2}$. We can factor out $\frac{1}{2}$ to get that: $$\sum_{k = 2}^\infty \frac{1}{2^{n-1}} = \frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{2^2}+\dots\right)$$ This part in the parenthesis is now in "standard form" (its initial term is $1$, and has some common ratio $r = \frac{1}{2}$). So, we can apply the geometric sum formula: $$1+r+r^2+\dots = \frac{1}{1-r},\quad\text{for }|r|<\frac{1}{2}$$ to get that: $$\sum_{k = 2}^\infty \frac{1}{2^{n-1}} = \frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{2^2}+\dots\right) = \frac{1}{2}\frac{1}{1-\frac{1}{2}} = \frac{1/2}{1/2} = 1$$

Mark Schultz-Wu
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Here's a proof without words picture

"proof without words" picture (Not sure how to make the picture smaller)

Nate 8
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1/2 +1/2^2 +1/2^3 ... = 1/2 +1/2(1/2+1/2^2+...) . That is it is a solution to x=1/2+1/2 x and x=1.

Jacob Wakem
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