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If you're looking for the asymptotic behavior, it may be of interest to approximate this process as a product of uniform distributions.

The product of $t$ uniform random variables has probability density

$$p_t(x) = \left|\frac{\log^{(t-1)} x}{(t-1)!}\right|$$

In particular

$$\int_{0}^{1/n} p_t(x)~\textrm{d}x = \frac{\Gamma(t,\log n)}{\Gamma(t)} = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n }{k!}$$

might be a decent estimate of $P(x_t=1)$

For instance, $P(x_2=1) \simeq \frac{\Gamma(2,\log n)}{\Gamma(2)} = \frac{\log n}{n} + \frac{1}{n}$... not quite $\frac{\log n}{n} + \frac{\gamma}{n} + O(\frac{1}{n^2})$ but not bad either.

I conjecture that the actual series is

$$P(x_t=1) = \frac{1}{n}\sum_{k=0}^{t-1} E_{t-1-k} \frac{\log^k n}{k!} + O\left(\frac{1}{n^2}\right)$$ with $$E_j = \frac{(-1)^{j}}{j!}\int_{0}^{\infty} e^{-x} \log^{j} x ~\textrm{d}x$$

If you're looking for the asymptotic behavior, it may be of interest to approximate this process as a product of uniform distributions.

The product of $t$ uniform random variables has probability density

$$p_t(x) = \left|\frac{\log^{(t-1)} x}{(t-1)!}\right|$$

In particular

$$\int_{0}^{1/n} p_t(x)~\textrm{d}x = \frac{\Gamma(t,\log n)}{\Gamma(t)} = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n }{k!}$$

might be a decent estimate of $P(x_t=1)$

For instance, $P(x_2=1) \simeq \frac{\Gamma(2,\log n)}{\Gamma(2)} = \frac{\log n}{n} + \frac{1}{n}$... not quite $\frac{\log n}{n} + \frac{\gamma}{n} + O(\frac{1}{n^2})$ but not bad either.

I conjecture (strongly) that the actual series is

$$P(x_t=1) = \frac{1}{n}\sum_{k=0}^{t-1} E_{t-1-k} \frac{\log^k n}{k!} + O\left(\frac{1}{n^2}\right)$$ with $$E_j = \frac{(-1)^{j}}{j!}\int_{0}^{\infty} e^{-x} \log^{j} x ~\textrm{d}x$$

If you're looking for the asymptotic behavior, it may be of interest to approximate this process as a product of uniform distributions.

The product of $t$ uniform random variables has probability density

$$p_t(x) = \left|\frac{\log^{(t-1)} x}{(t-1)!}\right|$$

In particular

$$\int_{0}^{1/n} p_t(x)~\textrm{d}x = \frac{\Gamma(t,\log n)}{\Gamma(t)} = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n }{k!}$$

might be a decent estimate of $P(x_t=1)$

For instance, $P(x_2=1) \simeq \frac{\Gamma(2,\log n)}{\Gamma(2)} = \frac{\log n}{n} + \frac{1}{n}$... not quite $\frac{\log n}{n} + \frac{\gamma}{n} + O(\frac{1}{n^2})$ but not bad either.

I conjecture that the actual series is

$$P(x_t=1) = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n}{k!} \left(\int_{0}^{\infty} \frac{(-1)^{(t-1-k)}}{(t-1-k)!}e^{-x} \log^{t-1-k} x ~\textrm{d}x\right) + O\left(\frac{1}{n^2}\right)$$

If you're looking for the asymptotic behavior, it may be of interest to approximate this process as a product of uniform distributions.

The product of $t$ uniform random variables has probability density

$$p_t(x) = \left|\frac{\log^{(t-1)} x}{(t-1)!}\right|$$

In particular

$$\int_{0}^{1/n} p_t(x)~\textrm{d}x = \frac{\Gamma(t,\log n)}{\Gamma(t)} = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n }{k!}$$

might be a decent estimate of $P(x_t=1)$

For instance, $P(x_2=1) \simeq \frac{\Gamma(2,\log n)}{\Gamma(2)} = \frac{\log n}{n} + \frac{1}{n}$... not quite $\frac{\log n}{n} + \frac{\gamma}{n} + O(\frac{1}{n^2})$ but not bad either.

I conjecture that the actual series is

$$P(x_t=1) = \frac{1}{n}\sum_{k=0}^{t-1} E_{t-1-k} \frac{\log^k n}{k!} + O\left(\frac{1}{n^2}\right)$$ with $$E_j = \frac{(-1)^{j}}{j!}\int_{0}^{\infty} e^{-x} \log^{j} x ~\textrm{d}x$$

If you're looking for the asymptotic behavior, it may be of interest to approximate this process as a product of uniform distributions.

The product of $t$ uniform random variables has probability density

$$p_t(x) = \left|\frac{\log^{(t-1)} x}{(t-1)!}\right|$$

In particular

$$\int_{0}^{1/n} p_t(x)~\textrm{d}x = \frac{\Gamma(t,\log n)}{\Gamma(t)} = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n }{k!}$$

might be a decent estimate of $P(x_t=1)$

For instance, $P(x_2=1) \simeq \frac{\Gamma(2,\log n)}{\Gamma(2)} = \frac{\log n}{n} + \frac{1}{n}$... not quite $\frac{\log n}{n} + \frac{\gamma}{n} + O(\frac{1}{n^2})$ but not bad either.

I conjecture that the actual series is

$$P(x_t=1) = \frac{1}{n}\sum_{k=0}^{t-1} \log^k n \left(\int_{0}^{\infty} \frac{(-1)^{(t-1-k)}}{(t-1-k)!}e^{-x} \log^{t-1-k} x ~\textrm{d}x\right) + O\left(\frac{1}{n^2}\right)$$

If you're looking for the asymptotic behavior, it may be of interest to approximate this process as a product of uniform distributions.

The product of $t$ uniform random variables has probability density

$$p_t(x) = \left|\frac{\log^{(t-1)} x}{(t-1)!}\right|$$

In particular

$$\int_{0}^{1/n} p_t(x)~\textrm{d}x = \frac{\Gamma(t,\log n)}{\Gamma(t)} = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n }{k!}$$

might be a decent estimate of $P(x_t=1)$

For instance, $P(x_2=1) \simeq \frac{\Gamma(2,\log n)}{\Gamma(2)} = \frac{\log n}{n} + \frac{1}{n}$... not quite $\frac{\log n}{n} + \frac{\gamma}{n} + O(\frac{1}{n^2})$ but not bad either.

I conjecture that the actual series is

$$P(x_t=1) = \frac{1}{n}\sum_{k=0}^{t-1} \frac{\log^k n}{k!} \left(\int_{0}^{\infty} \frac{(-1)^{(t-1-k)}}{(t-1-k)!}e^{-x} \log^{t-1-k} x ~\textrm{d}x\right) + O\left(\frac{1}{n^2}\right)$$