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Knowing the Stirling's approximation for the Gamma function (factorial) for integers: $$\Gamma(n+1)=n!\approx \sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$ Using the above approximation one can write: $$(n+1)!=\sqrt{2\pi(n+1)}(n+1)^{n+1}e^{-(n+1)}\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$ We know that following recursion holds: $$(n+1)!=(n+1)n!$$ One can rewrite this: $$(n+1)!=(n+1)\sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$ All this comes from: https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf (Page 8-9) Then the author gives this expansion to calculate the $a_k$ coefficients when $n$ becomes large. Comparing these two expressions for $(n+1)!$ gives $$1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots=\bigg(1+\frac{1}{n}\bigg)^{n+1/2}e^{-1}\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$ Then he says, that after "classical series expansion" this equals: $$1+\frac{a_1}{n}+\frac{a_2-a_1+\frac{1}{12}}{n^2}+\frac{\frac{13}{12}a_1-2a_2+a_3+\frac{1}{12}}{n^3}+\cdots$$ I don't understand how he got there. Only thing that came to my mind was, that as $n\to\infty$ $\big(1+\frac{1}{n}\big)^n$ goes to $e$ but then we are left with $\big(1+\frac{1}{n}\big)^{1/2}$. When i expand this into binomial series, i get: $$\bigg(1+\frac{1}{2n}-\frac{1}{8n^2}+\frac{1}{16n^3}\cdots\bigg)\cdot\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$ And I'm stuck here.

Or is there any other elementary way how to compute the coefficients $a_k$ of the stirling's series expansion for factorial/Gamma function?

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  • $\begingroup$ That part is easy, just Maclaurin expansion of $\sqrt{1+x}$. $\endgroup$ – Ian Jan 27 '18 at 19:43
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We expand the series up to terms of $\frac{1}{n^3}$. We obtain for $\left|\frac{1}{n}\right|<1$ \begin{align*} \color{blue}{\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}e^{-1}} &=e^{-1}e^{\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)}\\ &=\exp(-1)\exp\left[\left(n+\frac{1}{2}\right)\left(\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}+\cdots\right)\right]\\ &=\exp(-1)\exp\left(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\right)\\ &=\exp\left(\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\right)\\ &\color{blue}{=1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots}\tag{1} \end{align*}

and applying the binomial series expansion

we obtain \begin{align*} &\color{blue}{1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\frac{a_3}{(n+1)^3}+\cdots}\\ &\qquad =1+\frac{a_1}{n\left(1+\frac{1}{n}\right)}+\frac{a_2}{n^2\left(1+\frac{1}{n}\right)^2} +\frac{a_3}{n^3\left(1+\frac{1}{n}\right)^3}+\cdots\\ &\qquad =1+\frac{a_1}{n}\left(1-\frac{1}{n}+\frac{1}{n^2}\cdots\right) +\frac{a_2}{n^2}\left(1+\binom{-2}{1}\frac{1}{n}+\cdots\right) +\frac{a_3}{n^3}\left(1+\cdots\right)\\ &\qquad\color{blue}{=1+a_1\cdot\frac{1}{n}+\left(-a_1+a_2\right)\frac{1}{n^2} +\left(a_1-2a_2+a_3\right)\frac{1}{n^3}+\cdots}\tag{2}\\ \end{align*}

Putting (1) and (2) together

we obtain \begin{align*} &\left(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\right) \left(1+a_1\cdot\frac{1}{n}+\left(-a_1+a_2\right)\frac{1}{n^2} +\left(a_1-2a_2+a_3\right)\frac{1}{n^3}+\cdots\right)\\ &\quad\color{blue}{ =1+a_1\cdot\frac{1}{n}+\left(\frac{1}{12}-a_1+a_2\right)\frac{1}{n^2} +\left(-\frac{1}{12}+\frac{13}{12}a_1-2a_2+a_3\right)\frac{1}{n^3}+\cdots} \end{align*}

and the claim follows.

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  • $\begingroup$ Seems fine, just the last line of (1). How did you change $\exp\bigg(\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\bigg)$ into $\bigg(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\bigg)$ $\endgroup$ – Michal Dvořák Jan 30 '18 at 15:24
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    $\begingroup$ @MichalDvořák: We use the series expansion of $\exp(x)=1+x+\frac{x^2}{2}+\cdots$ here with $x=\frac{1}{12n^2}-\frac{1}{12n^3}$ and we only need to expand up to the linear term in $x$. $\endgroup$ – Markus Scheuer Jan 30 '18 at 15:27
  • $\begingroup$ @MichalDvořák: Many thanks for granting the bounty. $\endgroup$ – Markus Scheuer Jan 30 '18 at 20:32
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You were close. Note that1 $$ e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}=1+\frac{1}{12n^2}-\frac{1}{12n^3}+\frac{113}{1440n^4}+\cdots\tag1 $$

Therefore, you have $$ \bigg(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\bigg)\cdot\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\frac{a_3}{(n+1)^3}+\cdots\bigg)=\\ 1+\frac{a_1}{n}+\frac{a_2-a_1+\frac{1}{12}}{n^2}+\frac{a_3-2a_2+\frac{13}{12}a_1+\frac{1}{12}}{n^3}+\cdots\tag2 $$ as per $$ \frac{1}{n+1}=\frac{1}{n}-\frac{1}{n^2}+\cdots\\ \frac{1}{(n+1)^2}=\frac{1}{n^2}-\frac{2}{n^3}+\cdots\\ \frac{1}{(n+1)^3}=\frac{1}{n^3}-\frac{3}{n^4}+\cdots\\ \text{etc.}\tag3 $$


Easy (but slightly cumbersome) exercise: \begin{equation} \begin{aligned} {}&\lim_{n\to\infty}e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}=1\\ {}&\lim_{n\to\infty}n^2\left[e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}-1\right]=\frac{1}{12}\\ {}&\lim_{n\to\infty}n^3\left[e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}-1-\frac{1}{12n^2}\right]=-\frac{1}{12}\\ {}&\text{etc.} \end{aligned}\tag4 \end{equation}

These are basic limits that are left to the reader.

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  • $\begingroup$ Hang on, how did you come up with the first line? $\endgroup$ – Michal Dvořák Jan 29 '18 at 20:20
  • $\begingroup$ @MichalDvořák I edited in a hint. Let me know if you need some more details. Cheers! $\endgroup$ – AccidentalFourierTransform Jan 29 '18 at 20:27
  • $\begingroup$ I would be interested in deriving the (3) formulas, I'm unfamiliar with these, i got the (1) and the limits (4) $\endgroup$ – Michal Dvořák Jan 29 '18 at 20:35
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    $\begingroup$ @MichalDvořák Oh, it's pretty simple: $\frac{1}{n+1}=\frac{1}{n}\frac{1}{1+n^{-1}}=\frac{1}{n}(1-n^{-1}+n^{-2}-n^{-3}+\cdots)$ (just a geometric series in $n^{-1}$). Can you take it from here? $\endgroup$ – AccidentalFourierTransform Jan 29 '18 at 20:37
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    $\begingroup$ @MichalDvořák Instead of using the multinomial theorem, you can also use $\frac{1}{(n+1)^{a+1}}=\frac{(-1)^a}{a!}\left(\frac{\mathrm d}{\mathrm dn}\right)^a\frac{1}{n+1}$ and differentiate term-wise. $\endgroup$ – AccidentalFourierTransform Jan 29 '18 at 22:02

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