Stirling's Formula
[lim_{n
ightarrow infty} frac{n!}{(n^n/e^n)sqrt{2pi n}} =1.
]
Proof:
[egin{array}{ll}
n!
&= int_{0}^infty x^n e^{-x} mathrm{d}x \
&= int_{-sqrt{n}}^infty (n+sqrt{n}t)^n e^{-(n+sqrt{n}t)} sqrt{n} mathrm{d}t \
&= frac{n^n sqrt{n}}{e^n} int_{-sqrt{n}}^{infty} (1+frac{t}{sqrt{n}})^n e^{-sqrt{n}t}
mathrm{d}t. \
&= frac{n^n sqrt{n}}{e^n} int_{-infty}^{infty} f_n(t)
mathrm{d}t,
end{array}
]
其中
[f_{n}(t) =
left {
egin{array}{ll}
0 & t< sqrt{n} \
(1+frac{t}{sqrt{n}})^n e^{-sqrt{n}t} & tge sqrt{n}
end{array}
ight.
]
接下来证明(f_n(t))趋于(e^{-frac{t^2}{2}}),
[ln f_n(t) = n ln (1+ frac{t}{sqrt{n}}) - sqrt{n}t , t ge sqrt{n},
]
[ln (1+x) = 0 + x - frac{x^2}{2} + o(x^2),
]
当(n)足够大的时候
[ln f_n(t) = sqrt{n}t -t^2/2+sqrt{n}t+o(t^2/n)=-frac{t^2}{2}+o(t^2/n),
]
故(f_n(t)
ightarrow e^{-t^2/2}).
观察((t ge -sqrt{n}))
[egin{array}{ll}
frac{mathrm{d}}{mathrm{d}t}(ln f_{n+1}(t) - ln f_n(t) )
&= frac{sqrt{n}t}{sqrt{n}+t} - frac{sqrt{n+1}t}{sqrt{n+1}+t} \
&= frac{(sqrt{n}-sqrt{n+1})t^2}{(sqrt{n}+t)(sqrt{n+1}+t)} le 0,
end{array}
]
又(f_n(0)=0), 故
[f_{n+1} /f_n ge 1, quad t in [sqrt{n},0),
]
[f_{n+1} /f_n le 1, quad t in [0, +infty).
]
又(f_n(t))非负, 故根据单调收敛定理和优解控制定理可知
[lim_{n
ightarrow infty} int_{-infty}^{+infty} f_n(t) mathrm{d}t = int_{-infty}^{+infty} lim_{n
ightarrow infty} f_n(t) mathrm{d}t = int_{-infty}^{+infty}e^{-frac{t^2}{2}}mathrm{d} t=sqrt{2 pi}.
]
证毕.