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π
u
√ . Furthermore, since ue
is
√ +
where we have set s
2 2 6 The Coefficients of q(n) −u 2 27
√
∞ −u 2 π π
π
odd, it is equal to √ e du √ . Thus (21) formally
2 3 −∞ 2 3
becomes
√ √
e π 2n/3 2πèn n
q(n) ∼ √ . à 22)
n
4 3n e n!
And score another one for Stirling’s formula, which in turn
gives
√
e π 2n/3
q(n) ∼ √ , à 23)
4 3n
and our earlier estimate (18) allows us thereby to conclude that
√
e π 2n/3
p(n) ∼ √ . à 24)
4 3n

Success! We have determined the asymptotic formula for p(n)!
Well, almost. We still have two debts outstanding. We must justify
our formal passage to the limit in (21), and we must also prove our
evaluation (5). So first we observe that xe −x is maximized at x 1,
so we deduce that

s −s
√
1 + √ e n ≤ 1 (25)
n
s
(using x (1 + √ )) and also
n

s √
−s s 2
1 + √
e n ≤ e 2n (26)
n

2
s
(using x (1 + √ ) ).
n
Thus using (25) for positive s, by (21),
K n (s) ≤ e s 2 for s ≥ 0,