# беркай

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 Feb21 awarded Yearling Dec30 comment Evaluate $\sum_{n=1}^{\infty} \frac{\ln n}{(n+1)!}$ You may use Stirling's approximation. Dec28 asked Sources on Inverse Spectral Theory Dec6 comment Sufficient statistic for uniform distribution Check the Fisher–Neyman factorization theorem. Nov19 revised Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ added 1 characters in body Nov19 revised Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ deleted 1 characters in body Nov18 revised Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ added 50 characters in body Nov18 answered Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ Nov16 awarded Informed Nov16 revised Probability Question corrected spelling Nov16 suggested suggested edit on Probability Question Nov15 revised Find a Borel subset satisfying the condition corrected spelling and some math expressions Nov15 suggested suggested edit on Find a Borel subset satisfying the condition Nov12 revised Estimate the error of approximation of $ln⁡(1+x)$ deleted 29 characters in body Nov12 answered Estimate the error of approximation of $ln⁡(1+x)$ Jul24 answered Power series for $(1+x^3)^{-4}$ Jul23 comment Probability that none has to wait for change @blf Of course, it is a possiblity but if you place the first two places like this, you assume that the box office will gain 5 dollars in the third round. As far as I understand, the officer should be ready in each round to return money back. What if he receives 10 dollars in the third round in this case? Jul23 comment Probability that none has to wait for change @blf Again 5-dollar people cannot be in the positions 1 and 3. That's exactly why I first objected the use of Catalan numbers in this context. If you place 5-dollar people to 1 and 3.place, it means that one of the 10-dollar persons is on the 2.place which doesn't guarantee that he will return 5-dollars in the next round so this case cannot be considered. This is the way that I understood this problem, otherwise, your argument works! Jul23 comment Probability that none has to wait for change @blf That's good! I also want to understand the case $n=2$ which leads to $C_2=2$. Now, we should have in the first place 5-dollar person, second 5-dollar person, third and forth 10 dollar-persons. However, if we pick 5-dollar and 10-dollar persons identical then we have just 1 way to arrange them (This is not the case, indeed). Otherwise, assuming that they are different then we have 4 ways (5-dollar and 10 dollar-persons can change place among themselves). Most probably, you don't count the one who is the first or the last? What's the idea? Jul23 comment Probability that none has to wait for change @blf OK! Take $n=1$. If first 10-dollar person comes after first 5-dollar person then it fails because the box office now has 10 dollars (consisting of one 10 dollars and zero 5 dollars) and it cannot return 5 dollars in the next round if anyone pays 10 dollars. Do I misunderstand the question?