Question
An archive can restore backups numbered 1, 2, 3, ... in order of age, with no upper limit on the generation \(n\)umber. An audit selects generation \(n\) with probability proportional to \(\frac1{n(n+1)}\). The archive keeps only generations 1 through 8 on-site; an older request must be sent to off-site storage.
Find the constant that turns this rule into a probability distribution on all positive integers, then calculate the probability that an audit requests an off-site generation. What is the general probability of requesting a generation greater than \(k\)?
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Hint 1
Decompose \(\frac1{n(n+1)}\) into two adjacent reciprocals and add the first \(m\) terms.
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Solution
For every positive integer \(n\), \(\frac1{n(n+1)}=\frac1n-\frac1{n+1}\). Thus the finite partial sum is \(\sum_{n=1}^{m}1/[n(n+1)]=1-1/(m+1)\). Taking \(m\to\infty\) gives a total unnormalised mass of 1, so the required constant \(C\) satisfying \(C\sum\frac1{n(n+1)}=1\) is \(C=1\).
The first \(k\) generations carry probability \(1-\frac1{k+1}\). Their complement is therefore \(\Pr(N>k)=\frac1{k+1}\), for any integer \(k\ge0\). In particular, when only generations 1 through 8 are kept locally, the off-site-request probability is \(\Pr(N\ge9)=\Pr(N>8)=\frac19\approx11.11\%\). As a second calculation, the tail itself telescopes: \(\sum_{n=k+1}^{m}(1/n-1/(n+1))=1/(k+1)-1/(m+1)\), whose limit is \(\frac1{k+1}\). The law remains defined over infinitely many possible generations even though storage is finite.
Assumptions
The generation label is a positive integer without a finite maximum.
All probabilities refer to one audit request, and the weights are exactly proportional to \(\frac1{n(n+1)}\).
Off-site storage is needed precisely when \(N>8\).
Final Answer
The normalising constant is 1; \(\Pr(N>k)=\frac{1}{k+1}\), so \(\Pr(N>8)=\frac19\).