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Suppose you have the sequence of fractions $\left\{\frac{1}{a} : a \in \mathbb{N}\right\}$ ($\frac{1}{2},\frac{1}{3}$ and so on).

Now you start with $1$ and subtract every item of the sequence as long as the result is larger than $0$. You would start with subtracting $\frac{1}{2}$ and $\frac{1}{3}$, but then skip $4-6$ because the result would be 0 or smaller. You continue with $\frac{1}{7}$ and $\frac{1}{43}$.

Is there any lower limit to how small a number you can get or can you get as close to $0$ as you like?

Cameron Buie
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dtech
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    Check out this and read about its connection to Egyptian fractions. Since the sum of the reciprocals of the sequence approaches 1, your difference will approach 0. – Daniel R Nov 07 '13 at 15:40

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The limit is $0$, and you approach it fast (with few subtractions). Say after $k$ subtractions the remainder is $\dfrac{1}{n_k}$. Then the next number you subtract is $\dfrac{1}{n_k+1}$, and

$$\frac{1}{n_{k+1}} = \frac{1}{n_k} - \frac{1}{n_k+1} = \frac{1}{n_k(n_k+1)}.$$

The denominator satisfies $n_k \geqslant 2^{2^{k-1}}$ for $k \geqslant 1$, that is an extremely fast growing sequence.

Daniel Fischer
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    I interpret "and substract every item of the sequence as long as the result is larger than $0$" as saying that you do. You subtract the next $\frac1m$ that is smaller than the remainder. When the remainder is $\frac1{n_k}$, that is $\frac1{n_k+1}$. – Daniel Fischer Nov 07 '13 at 15:45
  • I apparently realised that seconds before you posted this and deleted my comment. You are correct – kaine Nov 07 '13 at 15:47
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There is no limit, you can get as close to 0 as you want.

The harmonic series diverge, which means you will never "run out" of fractions to subtract from, getting arbitrarily close to zero.

David
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    The fact that the harmonic series diverges is not sufficient to never "run out". – Did Nov 07 '13 at 15:43