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The following sum is (wrongly) obtained by trying a variation on Zeno's arrow paradox :

$$\left(1-\frac12\right) + \left(1-\frac12\right)\frac13+ \left(1-\frac12\right)\left(1-\frac13\right)\frac14+ \left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\frac15+\cdots$$

With following variation on Zeno's original problem : An arrow travelling between $0$ and $1$ , after travelling $\frac12$ distance, it will travel the $\frac13$ of remaining distance, after that $\frac14$ of the remaining distance, after that $\frac15$ of the remaining distance, and so on and so forth.

from the formulation this series has to converge to a value less than $\frac34$ .

My question is there a closed form known for this ? ( is the above sum correct formulation of the described problem ? No it does NOT)

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    $\begingroup$ A little calculation shows that your series is just $$\sum_{n=1}^\infty \frac{1}{n(n+1)}$$ which converges to $1$. So, may be your formulation is wrong. $\endgroup$ – Qurultay May 14 '19 at 9:35
  • $\begingroup$ @Qurultay : yes, you are correct, trying again $\endgroup$ – Arjang May 14 '19 at 9:39
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    $\begingroup$ Taking just the text part of your question, you argue that "from the formulation this series has to converge to a value less than $\frac34$ ". Why? $\endgroup$ – Andreas May 14 '19 at 9:55
  • $\begingroup$ @Andreas : It did make sense at the time, now that you asked I can't recall the logic I used. $\endgroup$ – Arjang May 14 '19 at 11:44
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    $\begingroup$ Thanks Arjang. I believe the mathematical problem formulation is actually a correct mapping of the text formulation, so the arrow indeed travels to position $1$. mihaild in his answer has given indications about the fractions such that it will not travel to position $1$. $\endgroup$ – Andreas May 14 '19 at 12:22
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I agree with Qurultay that the value is $1$. It's not hard to show that the sum can be written as

$$\frac12+\left(1-\frac12\right)\frac13+\left(1-\frac12\right)\left(1-\frac13\right)\frac14+\cdots=\sum_{n=2}^\infty\frac1n\prod_{k=2}^{n-1}\left(1-\frac1k\right)$$

The finite product is telescoping and reduces to $\frac1{n-1}$. Thus, we get a telescoping sum, namely

$$\sum_{n=2}^\infty\frac1n\prod_{k=2}^{n-1}\left(1-\frac1k\right)=\sum_{n=2}^\infty\frac1n\frac1{n-1}=\sum_{n=1}^\infty\frac1{n(n+1)}=1$$

$$\therefore~\frac12+\left(1-\frac12\right)\frac13+\left(1-\frac12\right)\left(1-\frac13\right)\frac14+\cdots~=~1$$

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In your variation, we have convergence to $1$. At each step the remaining part is multiplied by $\left(1 - \frac{1}{k}\right)$, so after $n$ steps we will left with $\prod\limits_{k=1}^n \left(1 - \frac{1}{k + 1}\right)$. As $\sum\limits_{k=1}^\infty \frac{1}{k + 1}$ diverges, this product goes to $0$.

To get convergence to something less than $1$ we need on each step to take part $a_n$ of what remains s.t. $\sum\limits_{n=1}^\infty a_n$ converges.

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