Josephus problem with an infinite sum, variant 2 (different sum)
This problem's question: What the infinite sum of the reciprocals of the size of circles,where circle sizes start at 1, where the survivor is the starter ?
A circle is formed of people, facing inward so the a person's left and right are defined. A person is picked to be the starter. You may consider them to be numbered from to , with number 1 being the starter. Then the process of removal from the circle begins. Every other person is removed from the circle going to the right. The first person removed is the person to the starter's right. In the case of a size 3 circle and using the proposed numbering: first person 2 is removed, then person 1 is removed and person 3 becomes the sole survivor. In the case of a size 1 starting circle, the starter is the survivor automatically.
The infinite sum is much simpler!
The answer is 2.
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1 solution
∑ n = 0 ∞ 2 − n ⇒ 2 . A proof by induction is sufficient to show that for powers of 2, position 1 is the survivor. To show that it is necessary, is slightly harder, the induction starts at 3, which is demonstrated in the problem statement as a case in which position 1 does not survive. Cases 1, 2 and 4 are powers of 2. After that, it is sufficient to notice that each time the circle cycles so that the next number that is smaller than the previous first position, position 1 is that that number. Therefore, only powers of 2 return position 1 to the front of the list.