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**mr fantastic** You have $\displaystyle \Pr(r) = \frac{r-1}{n} \sum_{j=r}^{n} \frac{1}{j-1} = \frac{r-1}{n} \left( \frac{1}{r-1} + \frac{1}{r} + \frac{1}{r+1} + .... + \frac{1}{n-1}\right)$.

If you draw a graph of $\displaystyle y = \frac{1}{u}$ and draw upper and lower rectangles over the appropriate partition it should be clear that

$\displaystyle \frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} + .... + \frac{1}{n-1} < \int_{r-1}^{n} \frac{1}{u} \, du < \frac{1}{r-1} + \frac{1}{r} + \frac{1}{r+1} + .... + \frac{1}{n-1}$

$\displaystyle \Rightarrow \frac{r-1}{n} \left( \frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} + .... + \frac{1}{n-1} \right) < \frac{r-1}{n} \int_{r-1}^{n} \frac{1}{u} \, du $

$\displaystyle < \frac{r-1}{n} \left( \frac{1}{r-1} + \frac{1}{r} + \frac{1}{r+1} + .... + \frac{1}{n-1} \right)$

Substitute $\displaystyle t = \frac{u}{n}$:

$\displaystyle \Rightarrow \frac{r-1}{n} \left( \frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} + .... + \frac{1}{n-1} \right) < \frac{r-1}{n} \int_{\frac{r-1}{n}}^{1} \frac{1}{t} \, dt $

$\displaystyle < \frac{r-1}{n} \left( \frac{1}{r-1} + \frac{1}{r} + \frac{1}{r+1} + .... + \frac{1}{n-1} \right)$.

Therefore, for large n and therefore large enough r: $\displaystyle \Pr(r) \approx x \int_{x}^{1} \frac{1}{t} \, dt$.

where the substitution $\displaystyle x = \frac{r-1}{n}$ is made.

Note that r-1 is the number of applicants initially rejected under the strategy and n is the total number of applicants. x therefore gives the proportion of applicants intially rejected.

All that's left is to find the value of x that maximises Pr(r) in this limit of large n ......