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Suppose I have something with length one unit. I divide it to two equal length $0.5$ unit and put left one in my left side and right one in my right side. I then do same for my right side and contact result's left one to my left side and replace my right side with it's right one.

Even if I do these infinitely, there's something at my right side remaining. So it seems that $$\frac12+\frac14+\frac18+\dots$$ never taste value $1$. Right?

NOT DUPLICATE: Actually answers at here does not answer my question. because if I accept them then I have $$\frac12+\frac14+\frac18+\dots=\frac23+\frac29+\frac2{27}+\dots=1$$ Then I can define $$A_n = \frac12+\frac14+\frac18+\dots+\frac1{2^n}$$ And $$B_n = \frac23+\frac29+\frac2{27}+\dots+\frac2{3^n}$$ Then, it's quite can be seen that: $$A_n<B_n , n\in\mathbb N$$

Then how could I accept that both series can reach each other at value $1$?!

Community
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Yasser Zamani
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    You should probably read http://en.wikipedia.org/wiki/Zeno's_paradoxes – Xoff Jan 24 '14 at 14:15
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    Good explanation here -http://acko.net/blog/to-infinity-and-beyond/ – GTX OC Jan 24 '14 at 14:16
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    Addition is a finite operation - there is no such thing as the sum of infinitely many terms until we define what that means. The best way to think about this case is to realize that if $a<b$ then there is a $c=\frac{a+b}2$ such that $a<c<b$. But if $a=\frac{1}{2}+\frac14\dots$ and $b=1$, can you find any $c$ so that $a<c<b$? – Thomas Andrews Jan 24 '14 at 15:20
  • @CameronBuie and other, I explined how this is different and is not duplicate with http://math.stackexchange.com/questions/149373/does-the-fact-that-sum-n-1-infty-1-2n-converges-to-1-mean-that-it-equal – Yasser Zamani Jan 24 '14 at 22:54
  • @ThomasAndrews, great! but then the summation of series infinitely is invalid while all places say it's equal to $1$. Also please read my edited question again where I explained how it's not duplicate, thanks! – Yasser Zamani Jan 24 '14 at 23:11
  • Did you skip the part where I said, "until we define what that means." The point is, you have to add a definition, the notion of a limit. @YasserZamani – Thomas Andrews Jan 24 '14 at 23:14
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    Why isn't that second sum zero? The first term is positive, and all the other terms are negative, and the limit of the sum is $0$. There's no obvious contradiction in saying it is zero. – Thomas Andrews Jan 24 '14 at 23:16
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    Note that $2/9 < 1/4$, since $8 < 9$; hence the second term (and all succeeding terms, actually) are negative. –  Jan 24 '14 at 23:20
  • @ThomasAndrews, I'm really sorry, it was my failure...I just tried to say, in first one series, what is remaining at my right side is greater than second one series which I added later. While if I accept both are $1$ then I have to accept no, both remaining at right side are equal. I think and will edit again with mathematic notations. – Yasser Zamani Jan 24 '14 at 23:28
  • @ThomasAndrews, I edited my question about how it's different with that question, thanks! – Yasser Zamani Jan 25 '14 at 19:48
  • That's the thing about limits. If you have $a_i<b_i$ you can only conclude that $\lim a_i \leq \lim b_i$. Consider just the case $B_n = A_{n+1}$. Then you want $B_n$ to converge to the same as $A_n$. @YasserZamani And you keep changing the question- the original question was a duplicate. – Thomas Andrews Jan 25 '14 at 23:49

5 Answers5

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It is true, that $$\sum_{k=1}^\infty \frac{1}{2^k} = 1$$ But only, if you take the whole, infinite series.

If you only take finitely many terms, let's say you have $\sum_{k=1}^n\frac{1}{2^k}$. This never equals $1$, but $1-\frac{1}{2^{n+1}}$.

So with only finitely many terms, you'll never reach $1$.

Alex R.
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  • Do you mean that exactly $0$ is at my right side at infinite? but I think it's $\frac{1}{2^\infty}$ is at my right side which does not exactly equal $0$! – Yasser Zamani Jan 24 '14 at 14:33
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    When we talk about an infinite series, we are talking about the limit of the partial sums. $\lim_{x \to \infty} \frac{1}{2^x}$ does equal zero, and that is what matters. – David Zhang Jan 24 '14 at 14:45
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    Wait, there's a typo in your answer: $\sum^{n}_{k=1}2^{-k}=1-2^{-n-1}$ (you wrote the RHS as $1-2^{-k-1}$). – Alex Nelson Jan 24 '14 at 15:41
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    oops, thanks, I'll correct it. – Alex R. Jan 24 '14 at 15:41
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I think the trouble here is that "equal" is a very strong word for what we're talking about. It's a question of being specific with definitions. The series you've effectively written down:

$$ \sum_{n=1}^\infty \frac{1}{2^n} $$

Is said to converge to 1, but not usually said to equal 1. This name much more accurately captures what you spoke of, that you never quite reach 1 after any finite number of "halvings", but that you clearly are heading that way.

This all comes from the fact that normally the definition of such an infinite series is the limit of the related sequence, so:

$$ \sum_{n=1}^\infty \frac{1}{2^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{1}{2^n} $$

unwitting
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Your "1/2+1/4+1/8+..." is by definition the limit of the sequence

$$S_n={1\over 2^1}+{1\over 2^2}+\cdots+{1\over 2^n}={2^n-1\over 2^n}$$

So, each $S_n<1$, but $\lim_{n\to\infty}S_n=1$.

Martín-Blas Pérez Pinilla
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Do you have any problem with $0.\overline{3} = \dfrac{1}{3}$? The left hand side really represents the sum $\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+ \dots$, and so the same problem exists here. The very nature of real numbers is that you can take limits of sequences of them. In some sense, all any real number is is a sequence of better and better approximations. This corresponds nicely with reality: you can never measure anything spot on, but you can continue to improve the sensitivity of you instruments to get better and better measurements.

Steven Gubkin
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I think the issue here is what we mean by 'is equal to' - does this even make sense for infinite series?

If we take "$a+b$ is a number" (the fact that we can add two numbers together) as an assumption, then we can define finite sums with any number of terms like this:

Assume $a_{1}+a_{2}+a_{3}+\ldots+a_{n-1}$ is a number. Then $$a_{1}+\ldots +a_{n-1}+a_{n}=(a_{1}+\ldots +a_{n-1})+a_{n}$$ is a number since this just uses our "$a+b$ is a number" assumption. This works for finite sums, but what do we do in the infinite case? Well, the natural definition is to say "What do the partial sums get close to?" - this isn't a perfectly precise definition, but it will work in this circumstance. We call this the 'limit' of the finite sums. In your case, you can see clearly that these partial sums get closer and closer to $1$, and if you follow our natural definition (and I stress again that this is not a perfect definition, but it does convey the feeling) then your infinite sum could be said to equal $1$.

In more technical contexts, different definitions of infinite summation make more sense, but you wouldn't encounter these very often.

preferred_anon
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