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Let's say you discover that you have to hold a cup of Diet Coke forever and if you spill it, you die. After the initial shock you will get better and better at not spilling. Let your chances of dying during the $t^{th}$ second be $D_t = \frac{1}{t+1}$. Can you survive for all time, assuming you would live forever if you didn't spill the Diet Coke?

The answer is, in this case, you can't. The chance you live forever is: $$\lim_{x\to\infty}\prod_{i=1}^x (1-D_i) = \lim_{x\to\infty}\prod_{i=1}^x \frac{i}{i+1} = \lim_{x\to\infty}\frac{1}{x+1} = 0$$

However, if you change $D_t$ to $\frac{1}{2^t}$ you can. The chances you die can be calculated like so: $$\frac{1}{2} + (1-\frac{1}{2})\frac{1}{4} + \frac{1}{2}(1-\frac{1}{4})\frac{1}{8} + \frac{1}{2}\frac{3}{4}(1-\frac{1}{8})\frac{1}{16} + \ldots$$ Which is less than $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots = 1$$

So therefore the chance you live forever is $\ge0$. The question is, what is the simplified probability of you being able to live forever?

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The probability you live forever is $\dfrac12 \times \dfrac34 \times \dfrac78 \times \cdots \approx 0.28878809508660242127889972192923078$

but OEIS A048651 does not suggest a simple closed form

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