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If we have a series starting at 1 and we keep adding half of the previous term and take an infinite amount of terms

$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...$$

I understand how we can say that the limit of the sum of this series approaches 2 (as I can make the sum as close to two as I want by taking at least $n$ number of terms), but is it correct to say that

$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...=2$$

And if yes, how can this be? After all, if I continue to keep taking half of the remaining distance between the sum and 2, even when the distance becomes infinitesimally small, I will still not arrive at 2...

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4 Answers 4

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In general, the notation

$$1 + \frac 1 2 + \frac 1 4 + ...$$

is used to denote the quantity

$$\lim_{n \to \infty} \sum\limits_{k = 1}^n \frac 1 {2^k}$$

which is, in fact, equal to $2$. Keep in mind that the limit really is the number $2$, not just approaching it; the sequence is approaching $2$.

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  • $\begingroup$ If I understand it correctly then, the limit is 2 (this part makes sense to me given the definition of the limit) but in actuality the sequence never makes it? I cannot say that the sum of this sequence is 2, as it never will be? However, I can say that the limit of the sum as the number of terms goes to infinity is 2? sum$\neq$limit of sum? $\endgroup$
    – dreamwalker
    Commented Sep 15, 2013 at 8:01
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    $\begingroup$ The "sum of the sequence" is the same as the "limit of the partial sums," which is $2$. $\endgroup$
    – user61527
    Commented Sep 15, 2013 at 8:07
  • $\begingroup$ Is this just by definition? I don't literally go out and add an infinite collection of terms (and if I somehow was able to do so I wouldn't get to 2 though I would get very close) but I fall back on the definition of the "sum of the sequence" and for all intents and purposes can say that it is equal to 2? $\endgroup$
    – dreamwalker
    Commented Sep 16, 2013 at 20:09
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You can't literally add an infinite collection of terms. What "the sum of the series" means is the limit of the partial sums as the number of terms goes to $\infty$.

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  • $\begingroup$ This is what powers that may be agreed to? If I understand it correctly then, I can't literally take an infinite collection of terms and add them up but by how the sum of the sequence is defined, I can figure out what the the sum approaches as number of terms goes to infinity and can rightfully say that that value (the limit) is the sum of the sequence? And by partial sums we mean sum with $n$ terms, then with $n+1$, $n+2$ and so on? Those would be partial sums? $\endgroup$
    – dreamwalker
    Commented Sep 15, 2013 at 17:26
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Here is another way of looking at it, you have the relationship $x = 1+\frac{1}{2} x$. Solving for $x$ gives $x=2$.

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    $\begingroup$ And I guess that, likewise, $1+2+4+8+16+\cdots=-1$... :-) $\endgroup$
    – Did
    Commented Sep 15, 2013 at 8:01
  • $\begingroup$ Who needs convergence :-). $\endgroup$
    – copper.hat
    Commented Sep 15, 2013 at 8:13
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    $\begingroup$ Right, only weak souls do... :-)) $\endgroup$
    – Did
    Commented Sep 15, 2013 at 11:36
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If you double the series, one can show that $2x-x=x$, by noting that it's

$$ 2+1+\frac 12 + \frac 14 + \frac 18 + \cdots - (1 + \frac 12 + \frac 14 + \frac 18 \cdots)$$

All of the terms disappear except the leading $2$.

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    $\begingroup$ This argument is already on the page. $\endgroup$
    – Did
    Commented Sep 15, 2013 at 11:36

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