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Stirling's formula $$N! \sim \sqrt{2 \pi}\ N^{N+ \frac{1}{2}} e^{-N}$$ follows easily from Laplace's method in light of the famous integral representation $$N! = \int_0^{\infty} e^{-z} z^N dz.$$ Basic representation theory of the symmetric group $S(N)$ gives the remarkable finite sum $$N! = \sum_{\lambda \in \mathbb{Y}_N} (\dim \lambda)^2$$ over all partitions of $N$, where $\dim \lambda$ counts the number of paths from $\emptyset$ to $\lambda$ in Young's lattice $\mathbb{Y}$.

  • Is it possible to derive Stirling's formula directly from this finite sum?

[Edit] I'd like to thank everyone for the very interesting comments below. For those looking at this thread for the first time, I was hoping that an answer to the question above might help us calculate the asymptotics of $$f(N,a) = \sum_{\lambda \in \mathbb{Y}_N} (\dim \lambda)^{a}.$$ Notice that the case $a=0$ is the partition function $|\mathbb{Y}_N|$.

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    $\begingroup$ Presumably one would have to find the largest term and how nearby terms deviate from this maximum. Results of this nature are due to Kerov, e.g., references [57][61][64] of math.mit.edu/~rstan/papers/ids.pdf. I don't know whether Kerov's work can lead to a proof of Stirling's formula. If so, it will be a much more difficult proof than the standard ones. $\endgroup$
    – Richard Stanley
    Commented Jul 18, 2011 at 3:15
  • $\begingroup$ Yes, it would certainly be more difficult, though it might help with the following. I'm interested in replacing the exponent "2" with an arbitrary integer a, which recovers the partition function at a=0. Computing the asymptotics in n of f(n,a) could hopefully interpolate between Stirling's formula and Hardy-Ramanujan's formula. However, it doesn't seem that such a function f(n,a) would yield a harmonic function / coherent system of probability measures in the sense of Kerov's asymptotic representation theory (the case a=2 for the Plancherel measure seems special in that regard). $\endgroup$
    – Alexander Moll
    Commented Jul 18, 2011 at 4:38
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    $\begingroup$ Replacing the exponent "2" with a complex variable results in what's known as the "Witten zeta function" or "representation zeta function" of the symmetric group $S_N$. This function has most combinatorial meaning at negative integer values for "2". See arxiv.org/abs/1102.4353 $\endgroup$
    – John Wiltshire-Gordon
    Commented Jul 18, 2011 at 5:00
  • $\begingroup$ It might be of interest to look into the work of Berele-Regev concerned with the asymptotics of Young tableaux in (k,l)-hooks (I think they deal with a question related to yours). Link: arxiv.org/PS_cache/arxiv/pdf/1007/1007.3833v1.pdf $\endgroup$
    – Vasu vineet
    Commented Jul 18, 2011 at 6:40
  • $\begingroup$ A related, though slightly different, question is addressed in a paper of A. Maroti (see MR2006609), where asymptotic estimates for the partition function are addressed via character theory of $S_n$. $\endgroup$
    – Geoff Robinson
    Commented Jul 18, 2011 at 7:13

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