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

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)

• 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. – Qurultay May 14 '19 at 9:35
• @Qurultay : yes, you are correct, trying again – Arjang May 14 '19 at 9:39
• 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? – Andreas May 14 '19 at 9:55
• @Andreas : It did make sense at the time, now that you asked I can't recall the logic I used. – Arjang May 14 '19 at 11:44
• 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 $1$. – Andreas May 14 '19 at 12:22

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

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

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

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.