NOTE & BLOG

    The Stirling Formula

    18 Sep, 2024

    Stirling's formula finds a relationship of n!,nⁿ,eⁿ while n trends to infinity.

    Wallis's Formula

    Let's start with a definite integral:

    ∫0π2sin⁡nx dx\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx

    We can derive the following result:

    ∫0π2sin⁡nx dx=−∫0π2sin⁡n−1x d(cos⁡x)\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = -\int_{0}^{\frac{\pi}{2}} \sin^{n-1} x \, d(\cos x)

    Using integration by parts, where ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du, we get:

    −∫0π2sin⁡n−1x d(cos⁡x)=−(sin⁡n−1xcos⁡x)∣0π2+∫0π2cos⁡x d(sin⁡n−1x)-\int_{0}^{\frac{\pi}{2}} \sin^{n-1} x \, d(\cos x) \\ = -\left( \sin^{n-1} x \cos x \right) \Big|_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \cos x \, d\left(\sin^{n-1} x \right) =∫0π2cos⁡2x⋅(n−1)sin⁡n−2x dx=(n−1)∫0π2(sin⁡n−2x−sin⁡nx) dx= \int_{0}^{\frac{\pi}{2}} \cos^2 x \cdot (n-1) \sin^{n-2} x \, dx \\ = (n-1) \int_{0}^{\frac{\pi}{2}} \left( \sin^{n-2} x - \sin^n x \right) \, dx

    Thus:

    ∫0π2sin⁡nx dx=(n−1)∫0π2(sin⁡n−2x−sin⁡nx) dx\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \left( \sin^{n-2} x - \sin^n x \right) \, dx

    We get:

    n∫0π2sin⁡nx dx=(n−1)∫0π2sin⁡n−2x dx  ⟹  ∫0π2sin⁡nx dx=n−1n∫0π2sin⁡n−2x dxn \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx \\ \implies \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx

    Therefore:

    For even nn, i.e., n=2kn = 2k:

    ∫0π2sin⁡2kx dx=2k−12k∫0π2sin⁡2k−2x dx=2k−12k⋅2k−32k−2⋯12∫0π2sin⁡0x dx=2k−12k⋅2k−32k−2⋯12⋅π2\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx = \frac{2k-1}{2k} \int_{0}^{\frac{\pi}{2}} \sin^{2k-2} x \, dx \\ = \frac{2k-1}{2k} \cdot \frac{2k-3}{2k-2} \cdots \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin^0 x \, dx \\= \frac{2k-1}{2k} \cdot \frac{2k-3}{2k-2} \cdots \frac{1}{2} \cdot \frac{\pi}{2}

    For odd nn, i.e., n=2k+1n = 2k + 1:

    ∫0π2sin⁡2k+1x dx=2k2k+1∫0π2sin⁡2k−1x dx=2k2k+1⋅2k−22k−1⋯23∫0π2sin⁡1x dx=2k2k+1⋅2k−22k−1⋯23\int_{0}^{\frac{\pi}{2}} \sin^{2k+1} x \, dx \\ = \frac{2k}{2k+1} \int_{0}^{\frac{\pi}{2}} \sin^{2k-1} x \, dx \\ = \frac{2k}{2k+1} \cdot \frac{2k-2}{2k-1} \cdots \frac{2}{3} \int_{0}^{\frac{\pi}{2}} \sin^1 x \, dx \\ = \frac{2k}{2k+1} \cdot \frac{2k-2}{2k-1} \cdots \frac{2}{3}

    Here, we need the double factorial:

    m!!=m(m−2)(m−4)⋯m!! = m (m-2) (m-4) \cdots

    Thus:

    ∫0π2sin⁡2kx dx=(2k−1)!!(2k)!!⋅π2\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2} ∫0π2sin⁡2k+1x dx=(2k)!!(2k+1)!!\int_{0}^{\frac{\pi}{2}} \sin^{2k+1} x \, dx = \frac{(2k)!!}{(2k+1)!!}

    Since sin⁡2k+1x<sin⁡2kx<sin⁡2k−1x\sin^{2k+1} x < \sin^{2k} x < \sin^{2k-1} x for x∈(0,π2)x \in (0, \frac{\pi}{2}), we have:

    ∫0π2sin⁡2k+1x dx<∫0π2sin⁡2kx dx<∫0π2sin⁡2k−1x dx\int_{0}^{\frac{\pi}{2}} \sin^{2k+1} x \, dx < \int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx < \int_{0}^{\frac{\pi}{2}} \sin^{2k-1} x \, dx

    Substituting the above formulas:

    (2k)!!(2k+1)!!<(2k−1)!!(2k)!!⋅π2<(2k−2)!!(2k−1)!!\frac{(2k)!!}{(2k+1)!!} < \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2} < \frac{(2k-2)!!}{(2k-1)!!}

    We can examine the difference between the bounds for ∫0π2sin⁡2kx dx\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx:

    lim⁡k→∞((2k−2)!!(2k−1)!!−(2k)!!(2k+1)!!)=lim⁡k→∞((2k−2)!!(2k−1)!!−2k2k+1⋅(2k−2)!!(2k−1)!!)=lim⁡k→∞12k+1⋅(2k−2)!!(2k−1)!!=0\lim_{k \rightarrow \infty} \left( \frac{(2k-2)!!}{(2k-1)!!} - \frac{(2k)!!}{(2k+1)!!} \right) \\ = \lim_{k \rightarrow \infty} \left( \frac{(2k-2)!!}{(2k-1)!!} - \frac{2k}{2k+1} \cdot \frac{(2k-2)!!}{(2k-1)!!} \right) \\ = \lim_{k \rightarrow \infty} \frac{1}{2k+1} \cdot \frac{(2k-2)!!}{(2k-1)!!} = 0

    Note: Here 2k−22k−1<1\frac{2k-2}{2k-1} < 1

    For the inequality:

    (2k)!!(2k+1)!!<(2k−1)!!(2k)!!⋅π2<(2k−2)!!(2k−1)!!\frac{(2k)!!}{(2k+1)!!} < \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2} < \frac{(2k-2)!!}{(2k-1)!!}

    Multiply all terms by (2k)!!(2k−1)!!\frac{(2k)!!}{(2k-1)!!}:

    (2k)!!2(2k+1)!!(2k−1)!!<π2<(2k−2)!!(2k)!!(2k−1)!!2\frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} < \frac{\pi}{2} < \frac{(2k-2)!! (2k)!!}{(2k-1)!!^2}

    We can further examine the difference between the bounds for π2\frac{\pi}{2}:

    lim⁡k→∞((2k−2)!!(2k)!!(2k−1)!!2−(2k)!!2(2k+1)!!(2k−1)!!)=lim⁡k→∞(12k−12k+1)⋅(2k)!!2(2k+1)!!(2k−1)!!=lim⁡k→∞12k⋅(2k)!!2(2k+1)!!2=0\lim_{k \rightarrow \infty} \left( \frac{(2k-2)!! (2k)!!}{(2k-1)!!^2} - \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} \right) \\= \lim_{k \rightarrow \infty} \left( \frac{1}{2k} - \frac{1}{2k+1} \right) \cdot \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} \\= \lim_{k \rightarrow \infty} \frac{1}{2k} \cdot \frac{(2k)!!^2}{(2k+1)!!^2} = 0

    Thus, we obtain Wallis's Formula:

    lim⁡k→∞(2k)!!2(2k+1)!!(2k−1)!!=π2\lim_{k \rightarrow \infty} \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} = \frac{\pi}{2}

    This formula introduces Wallis's result in preparation for deriving Stirling's formula.

    Stirling's Formula

    Stirling's formula reveals a relationship between n!n!, nnn^n, and ene^n.

    As nn approaches infinity, n!n! is approximately 2πn(ne)n\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n.

    First, introduce a limit:

    lim⁡n→∞(n!nn)1n=1e\lim_{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} = \frac{1}{e}

    Proof:

    lim⁡n→∞(n!nn)1n=lim⁡n→∞e1nln⁡(n!nn)\lim_{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} = \lim_{n \rightarrow \infty} e^{\frac{1}{n} \ln \left( \frac{n!}{n^n} \right)}

    which is equivalent to proving:

    lim⁡n→∞1nln⁡(n!nn)=−1\lim_{n \rightarrow \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) = -1

    We have:

    lim⁡n→∞1nln⁡(n!nn)=lim⁡n→∞1n(ln⁡1n+ln⁡2n+⋯+ln⁡nn)=∫01ln⁡x dx=(xln⁡x−x)∣01=−1\lim_{n \rightarrow \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) \\ = \lim_{n \rightarrow \infty} \frac{1}{n} \left( \ln \frac{1}{n} + \ln \frac{2}{n} + \cdots + \ln \frac{n}{n} \right) \\ = \int_{0}^{1} \ln x \, dx = (x \ln x - x) \Big|_{0}^{1} = -1

    QED.

    Thus,

    n!nn∼1en\frac{n!}{n^n} \sim \frac{1}{e^n}

    We would like to verify:

    lim⁡n→∞n!nnen=1\lim_{n \rightarrow \infty} \frac{n!}{n^n} e^n = 1

    Let’s test this.

    Using the ratio test:

    an=n!nnen,an+1=(n+1)!(n+1)n+1en+1a_n = \frac{n!}{n^n} e^n, \quad a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}} e^{n+1} lim⁡n→∞an+1an=e(1+1n)n>1\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^n} > 1

    Although the limit of the expression is ee, it is always approaching ee from below, so it is still greater than 1, indicating that ana_n is divergent.

    Mathematicians have attempted various approaches and found:

    an=n!nn+12ena_n = \frac{n!}{n^{n + \frac{1}{2}}} e^n

    To prove that ana_n converges, use the ratio test again:

    an=n!nn+12en,an+1=(n+1)!(n+1)n+32en+1a_n = \frac{n!}{n^{n + \frac{1}{2}}} e^n, \quad a_{n+1} = \frac{(n+1)!}{(n+1)^{n + \frac{3}{2}}} e^{n+1} lim⁡n→∞an+1an=e(1+1n)n+12\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}}}

    Compare this with:

    (1+1n)n+12=e(n+12)ln⁡(1+1n)\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}} = e^{(n + \frac{1}{2}) \ln \left(1 + \frac{1}{n}\right)}

    and ee.

    Thus, compare:

    (n+12)ln⁡(1+1n)(n + \frac{1}{2}) \ln \left(1 + \frac{1}{n}\right)

    with 1.

    Specifically, compare ln⁡n+1n\ln \frac{n+1}{n} with 1n+12\frac{1}{n + \frac{1}{2}}:

    Since:

    1n+1<ln⁡(1+1n)<1n\frac{1}{n + 1} < \ln \left(1 + \frac{1}{n}\right) < \frac{1}{n}

    Note: The right side follows from the inequality ln⁡(1+x)<x\ln(1 + x) < x (for x≠0x \neq 0).

    The left side follows from: 1−nn+1<ln⁡(n+1n)1 - \frac{n}{n+1} < \ln \left(\frac{n+1}{n}\right), where nn+1\frac{n}{n+1} is considered as a whole. This is high school knowledge.

    Because 1x\frac{1}{x} is a concave function:

    ∫nn+11x dx=ln⁡n+1n>1n+12\int_{n}^{n+1} \frac{1}{x} \, dx = \ln \frac{n+1}{n} > \frac{1}{n + \frac{1}{2}}

    Therefore:

    lim⁡n→∞an+1an=e(1+1n)n+12<1\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}}} < 1

    Thus, ana_n is monotonically decreasing and an>0a_n > 0, so it must converge, i.e., it has a limit.

    Let:

    lim⁡n→∞n!nn+12en=a\lim_{n \rightarrow \infty} \frac{n!}{n^{n + \frac{1}{2}}} e^n = a

    Then, we need to find this limit.

    From the earlier Wallis formula:

    lim⁡k→∞(2k)!!2(2k+1)!!(2k−1)!!=π2\lim_{k \rightarrow \infty} \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} = \frac{\pi}{2} (2n)!!=2n(2n−2)⋯2=2nn!(2n)!! = 2n (2n-2) \cdots 2 = 2^n n! (2n−1)!!=(2n)!(2n)!!=(2n)!2nn!(2n-1)!! = \frac{(2n)!}{(2n)!!} = \frac{(2n)!}{2^n n!} (2n+1)!!=(2n+2)!(2n+2)!!=(2n+2)!2n+1(n+1)!(2n+1)!! = \frac{(2n+2)!}{(2n+2)!!} = \frac{(2n+2)!}{2^{n+1} (n+1)!}

    Thus, Wallis' formula becomes:

    lim⁡n→∞16n(n!)4(2n+2)(2n)!(2n+2)!=π2\lim_{n \rightarrow \infty} \frac{16^n (n!)^4 (2n+2)}{(2n)!(2n+2)!} = \frac{\pi}{2}

    Using:

    n!∼ae−nnn+12n! \sim a e^{-n} n^{n + \frac{1}{2}}

    Substitute into:

    lim⁡n→∞a2e24(n+1n)2n+32=π2\lim_{n \rightarrow \infty} \frac{a^2 e^2}{4 \left(\frac{n+1}{n}\right)^{2n + \frac{3}{2}}} = \frac{\pi}{2}

    Solve for:

    a=2πa = \sqrt{2 \pi}

    Thus, as nn approaches infinity:

    n!∼2πe−nnn+12n! \sim \sqrt{2 \pi} e^{-n} n^{n + \frac{1}{2}}

    or equivalently:

    n!∼2πn(ne)nn! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n
    # MATHEMATICS