12 1. Preliminary Notions which can easily be proved by observing that (1.13) log(n!) = log 2 + · · · + log n 2 1 log t dt + 3 2 log t dt + · · · + n n−1 log t dt = n 1 log t dt = t log t − t t=n t=1 = n log n − n + 1 log n e n . We now state Stirling’s formula in its traditional form. Theorem 1.14. (Stirling’s formula) As n → ∞, (1.14) n! ∼ nne−n √ 2πn. Proof. First observe that since the function log t is increasing for t 0, it is clear that j j−1 log t dt log j j+1 j log t dt. Adding up these inequalities, for j = 1, 2,...,n, we obtain n 0 log t dt log n! n+1 1 log t dt, n log n − n log n! (n + 1) log(n + 1) − n. Rearranging this last expression, we obtain n log n − n log n! n log n − n + log n + O(1), suggesting that one should consider the expression bn := n log n − n + 1 2 log n as an approximation of log n!. So, let us consider the difference an := log n! − bn = log n! − n + 1 2 log n + n. Observe that, setting α = 1/(2n + 1) and using the fact that 1 2 log 1 + α 1 − α = α + 1 3 α3 + 1 5 α5 + · · · , we get that 0 an − an+1 = n + 1 2 log n + 1 n − 1 = 1 α · 1 2 · log 1 + α 1 − α − 1

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