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

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

A: 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.
