Using /u/ray_giraffe's idea, it's possible to show rigorously that P(n) ~ nlog2. The 3/2 may be a bit trickier.

Notice that your equation is equivalent to

[;1 = \sum_{k=1}^{n-1} \left(\frac{k}{n}\right)^x = \sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^x\right) ;]

We want to say that [;\lim_{n \to \infty} \sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^{n\log 2} = \sum_{k=1}^\infty \lim_{n \to \infty} \left(1-\frac{k}{n}\right)^{n\log 2} = \sum_{k=1}^\infty 2^{-k} = 1;]

To do this, consider the family of sequences an^k defined by

[; a_n^k = \left(1-\frac{k}{n}\right)^{n\log 2} ;] for 0 < k < n and 0 otherwise.

Notice that since (1-k/n) and nlog 2, both considered as functions of n, is increasing for n > k, for fixed k a_n^k is increasing as a function of n. Since

[;\lim_{n \to \infty} a_n^k = 2^{-k}, ;] we have that an^k <= 2^-k for all n. Since this sequence is summable, the dominated convergence theorem lets us exchange the limits as we wanted to above, since of course

[;\sum_{k=1}^\infty a_n^k = \sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^{n\log 2}.;]

In particular, for ε > 0 and n large,

[; 1-\sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^{n\log 2}< \varepsilon,;]

since the second summand is less than 1 for the same reason as above.

Plugging in the definition of P(n), we have equivalently that

[;\sum_{k=1}^{n-1}\left(1-\frac{k}{n}\right)^{P(n)}-\left(1-\frac{k}{n}\right)^{n\log 2} < \varepsilon;]

Since each term must have the same sign (0 <1-k/n < 1), it follows that each term is positive and so for 0 < k < n [;0 < \left(1-\frac{k}{n}\right)^{P(n)} - \left(1-\frac{1}{n}\right)^{n\log2} < \varepsilon,;]

In particular (setting k = n-k), we have that

[;\lim_{n \to \infty} \left(\frac{k}{n}\right)^{P(n)} = \left(\frac{k}{n}\right)^{n\log 2} = 2^{-k}.;]

To show that P(n)/n->log 2, it suffices to show that P(n) -nlog 2 is bounded. Suppose not, and for each m, there is some n_m with P(n)-nlog2 > m. Then for all 0 < k < n_m

[; \left(\frac{k}{n_m}\right)^{P(n_m)-n-M\log 2} < \left(\frac{k}{n_m}\right)^m < 1;]

(the second inequality follows since k < n_m).

Taking m to infinity shows that

[; 1 \leq \lim_{m \to \infty}\left(\frac{k}{n_m}\right)^m \leq 1,;]

i.e.

[; \lim_{n \to \infty} \left(\frac{k}{m_n}\right)^m = 1;]

Since this is true for all k, in particular

[;\lim_{n \to \infty} 2^m = \lim_{n \to \infty} \left(\frac{2}{m_n}\right)^m\frac{1}{\left(\frac{1}{m_n}\right)^m = 1,;]

an obvious contradiction.