Exercise 8.30

Question

Exercise 30: 
    Use Stirling's formula to prove that $$\lim_{x\rightarrow\infty}\frac{\Gamma(x+c)}{x^c\Gamma(x)}=1$$ for every real constant $c$.

analambanomenos · Accepted Answer

From Stirling's formula we get the limits $$\lim_{xightarrow\infty}\frac{x^c\Gamma(x)}{x^c\big(\frac{x-1}{e}\big)^{x-1}\sqrt{2\pi(x-1)}}=1\quad\quad\lim_{xightarrow\infty}\frac{\Gamma(x+c)}
    {\big(\frac{x+c-1}{e}\big)^{x+c-1}\sqrt{2\pi(x+c-1)}}=1.$$ Hence 
    \begin{align*}
        1 &= \lim_{xightarrow\infty}\frac{\Gamma(x+c)}{x^c\Gamma(x)}\cdot\frac{x^c\big(\frac{x-1}{e}\big)^{x-1}\sqrt{2\pi(x-1)}}{\big(\frac{x+c-1}{e}\big)^{x+c-1}\sqrt{2\pi(x+c-1)}} \
        &= \lim_{xightarrow\infty}\frac{\Gamma(x+c)}{x^c\Gamma(x)}\cdot\frac{x^ce^c(x-1)^{x-1}}{(x+c-1)^{x+c-1}}\sqrt{1-\frac{c}{x+c-1}} \
        &= \lim_{xightarrow\infty}\frac{\Gamma(x+c)}{x^c\Gamma(x)}\cdot\frac{x^ce^c(x-1)^{x-1}}{(x+c-1)^{x+c-1}}.
    \end{align*}
    So to show that the limit of the first factor is 1, it suffices to show that the limit of the second factor is also 1.

By substituting $y$ for $x-1$, we have
    \begin{align*}
        \lim_{xightarrow\infty}\frac{x^ce^c(x-1)^{x-1}}{(x+c-1)^{x+c-1}} &= \lim_{yightarrow\infty}\frac{(y+1)^ce^cy^y}{(y+c)^{y+c}} \
        &= \lim_{yightarrow\infty}\bigg(1+\frac{1-c}{y+c}\bigg)^c\cdot\frac{e^cy^y}{(y+c)^y} \
        &= \lim_{yightarrow\infty}\frac{e^c}{(1+c/y)^y} \
        &= 1
    \end{align*}
    The last equality comes from Exercise 4(d).

Exercise 8.30

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Exercise 8.30

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