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It doesn't seem straightforward to put this into mathematical notation, but I'll do my best to explain the setup. Consider a harmonic series of the following type. For the sake of argument, say we have a set of 5 elements which are the reciprocals of the first 5 natural numbers:

$\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\right\}$

Start the series index at k = 1, and define the first element of the series to be 1. Knock 1 out of the set, so we now have:

$\left\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\right\}$

For the second summand in the series, index k = 2, make a uniform random sample from the above set. Say we come up with $\frac{1}{4}$. Our series is now:

$1 + \frac{1}{4}$

Knock $\frac{1}{2}$ out of the set, as its denominator corresponds to the current index k = 2. We now have the set:

$\left\{\frac{1}{3}, \frac{1}{4}, \frac{1}{5}\right\}$

For the third summand in the series, again choose randomly from the above set, and knock out the value associated with the current index afterwards, that is, $\frac{1}{3}$.

For any given number k, i.e. size of both the series and starting size of the set, there should be an expected value associated with the stochastic series; using linearity of expectation it is straightforward to calculate for k = 5 as

$1 + \frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}}{4} + \frac{\frac{1}{3} + \frac{1}{4} + \frac{1}{5}}{3} + \frac{ \frac{1}{4} + \frac{1}{5}}{2} + \frac{1}{5}.$

I'm interested in what happens in the limit as k goes to infinity in this process. I've done some experiments numerically, and it seems that the expected value for large k converges to a number, $\approx 1.64$. Can it be proved that the expected value converges in the limit $k \to \infty$, and can the exact value be found?

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up vote 3 down vote accepted

Apart from the $1$ which doesn't follow the rule, in the expected value you seem to have terms $$\dfrac{1}{k+1-j} \sum_{i=j}^k \dfrac{1}{i}$$ for $j$ from $2$ to $k$. I don't know if there's a closed form for the sum of that, but it appears to have generating function $$ G(t) = \left(\frac{\pi^2}{6} + \mathrm{dilog}(t) - \ln(t)\ln(1-t)\right) \frac{t}{1-t} -\frac{\ln(1-t)^2}{2}$$

As for asymptotics, since $\displaystyle\sum_{i=j}^k 1/i \approx \int_{j}^{k+1} ds/s = \ln(k+1)-\ln(j)$, and $$\displaystyle\int_{2}^k \dfrac{\ln(k+1)-\ln(s)}{k+1-s} \ ds = \mathrm{dilog}\left(\frac{2}{k+1}\right) - \mathrm{dilog}\left(\frac{k}{k+1}\right) \to \frac{\pi^2}{6}\ \text{ as } k \to \infty$$ I think the limit is $\pi^2/6$.

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Wow, the stochastic Basel problem! – Bitrex Jul 11 '13 at 3:25

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