Puiseux Expansion of Gamma Function about Infinity

Question

In trying to find interesting proofs that Student's T Distribution converges to the Regularized Normal Distribution when $k$ (the number of desgrees of freedom) grows without bounds (i.e. $= \infty$).
One of the ways I tried involved trying to find an expansion for $\frac{\Gamma\left(\frac{v+1}{2}\right)}{\Gamma\left(\frac{v}{2}\right)}$ about $v=\infty$, though I could not make any headway on finding the expansion and I feared it would be extremely complicated. However, when I asked Wolfram Alpha I got a beautiful Puiseux Series, which is expressed as: $$\frac{\Gamma\left(\frac{v+1}{2}\right)}{\Gamma\left(\frac{v}{2}\right)}\simeq \frac{v^{1/2}}{\sqrt2}-\frac{v^{-1/2}}{4\sqrt2}+\frac{v^{-3/2}}{32\sqrt2}+5\frac{v^{-5/2}}{128\sqrt2}-21\frac{v^{-5/2}}{2048\sqrt2}+\cdots$$ (more terms can be found in the link provided if desired). I can't help but notice that the denominators are simply $2^n \sqrt{2}$ for $n\in(0,2,5,7,11,\ldots)$ However, I can't find any pattern in $n$, nor can I explain the additional coefficients that appear starting on the fourth term, so I can't think of a way to work backward to find a proof.
Regardless, all I need is the first term for the purpose of taking a limit, as all the other terms will vanish anyway. Does anyone know how to prove the expansion?

Edit: I should note that I have already attacked the problem using Stirlings Series, but I found that to be somewhat brute force for such a simple summation and I hoped for a more clever argument (it does however some light on where the coefficients come from). Again, I only need to know the series up to $\frac{v^{1/2}}{\sqrt2} + O(v^{-1/2})$

Answer 1

Powers of $\boldsymbol{2}$ in the denominator

This is only an observational guess, but I have computed the asymptotic series for $$ \frac{\Gamma\left(x+\frac12\right)}{\Gamma(x)\sqrt{x}} $$ out a large number of terms and the exponents of $2$ in the denominators seem to be $$ 3n+\sum_{k=1}^\infty\left\lfloor\frac{n}{2^k}\right\rfloor $$ Thus, the exponents of $2$ in the denominators of the asymptotic series for $$ \frac{\Gamma\left(\frac{v+1}2\right)}{\Gamma\left(\frac{v}2\right)\sqrt{\frac{v}2}} $$ would be $$ \bbox[5px,border:2px solid #C0A000]{2n+\sum_{k=1}^\infty\left\lfloor\frac{n}{2^k}\right\rfloor} $$ This gives denominators of $$ 2^0,2^2,2^5,2^7,2^{11},2^{13},2^{16},2^{18},2^{23},2^{25},2^{28},\dots $$

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The Asymptotic Series

To compute the asymptotic series, we can compute the series at infinity $$ \begin{align} &\textstyle\log\left(x-\tfrac12\right)-\log(x-1)\\ &\textstyle=\log\left(1-\tfrac1{2x}\right)-\log\left(1-\tfrac1x\right)\\ &\textstyle=\frac1{2x}+\frac3{8x^2}+\frac7{24x^3}+\frac{15}{64x^4}+\frac{31}{160x^5}+\frac{21}{128x^6}+\frac{127}{896x^7}+\frac{255}{2048x^8}+\frac{511}{4608x^9}+\frac{1023}{10240x^{10}}\\ &\textstyle+\frac{2047}{22528x^{11}}+\frac{1365}{16384x^{12}}+\frac{8191}{106496x^{13}}+\frac{16383}{229376x^{14}}+\frac{32767}{491520x^{15}}+\frac{65535}{1048576x^{16}}+\frac{131071}{2228224x^{17}}\\ &\textstyle+\frac{29127}{524288x^{18}}+\frac{524287}{9961472x^{19}}+\frac{209715}{4194304x^{20}}+\frac{299593}{6291456x^{21}}+\frac{4194303}{92274688x^{22}}+\frac{8388607}{192937984x^{23}}+O\!\left(\frac1{x^{24}}\right)\tag{1} \end{align} $$ Then we can apply the Euler-Maclaurin Sum Formula to get $$ \begin{align} &\textstyle\log\left(\frac{\large\Gamma\left(x+\frac12\right)}{\large\Gamma\left(x\right)\vphantom{\frac12}}\right)\\ &\textstyle=\frac12\log(x)-\frac1{8x}+\frac1{192x^3}-\frac1{640x^5}+\frac{17}{14336x^7}-\frac{31}{18432 x^9}+\frac{691}{180224x^{11}}-\frac{5461}{425984x^{13}}\\ &\textstyle+\frac{929569}{15728640x^{15}}-\frac{3202291}{8912896x^{17}}+\frac{221930581}{79691776x^{19}}-\frac{4722116521}{176160768x^{21}}+O\!\left(\frac1{x^{23}}\right)\tag{2} \end{align} $$ Then we can exponentiate to get $$ \begin{align} &\textstyle\frac{\large\Gamma\left(x+\frac12\right)}{\large\Gamma\left(x\right)\vphantom{\frac12}\sqrt{x}}\\ &\textstyle=1-\frac1{8x}+\frac1{128x^2}+\frac5{1024x^3}-\frac{21}{32768x^4}-\frac{399}{262144x^5}+\frac{869}{4194304x^6}+\frac{39325}{33554432x^7}\\ &\textstyle-\frac{334477}{2147483648x^8}-\frac{28717403}{17179869184x^9}+\frac{59697183}{274877906944x^{10}}+\frac{8400372435}{2199023255552x^{11}}-\frac{34429291905}{70368744177664x^{12}}\\ &\textstyle-\frac{7199255611995}{562949953421312x^{13}}+\frac{14631594576045}{9007199254740992x^{14}}+\frac{4251206967062925}{72057594037927936x^{15}}-\frac{68787420596367165}{9223372036854775808x^{16}}\\ &\textstyle-\frac{26475975382085110035}{73786976294838206464x^{17}}+\frac{53392138323683746235}{1180591620717411303424x^{18}}+\frac{26275374869163335461975}{9444732965739290427392x^{19}}\\ &\textstyle-\frac{105772979046693606062363}{302231454903657293676544x^{20}}-\frac{64759060397977041632277937}{2417851639229258349412352x^{21}}+\frac{130175508110590141461339987}{38685626227668133590597632x^{22}}\\ &\textstyle+O\!\left(\frac1{x^{23}}\right)\tag{3} \end{align} $$ Then we can substitute $x=\frac v2$ to get $$ \begin{align} &\textstyle\frac{\large\Gamma\left(\frac{v+1}2\right)}{\large\Gamma\left(\frac{v}2\right)\sqrt{\frac{v}2}}\\ &\textstyle=1-\frac1{4v}+\frac1{32v^2}+\frac5{128v^3}-\frac{21}{2048v^4}-\frac{399}{8192v^5}+\frac{869}{65536v^6}+\frac{39325}{262144v^7}\\ &\textstyle-\frac{334477}{8388608v^8}-\frac{28717403}{33554432x^9}+\frac{59697183}{268435456v^{10}}+\frac{8400372435}{1073741824v^{11}}-\frac{34429291905}{17179869184v^{12}}\\ &\textstyle-\frac{7199255611995}{68719476736v^{13}}+\frac{14631594576045}{549755813888v^{14}}+\frac{4251206967062925}{2199023255552v^{15}}-\frac{68787420596367165}{140737488355328v^{16}}\\ &\textstyle-\frac{26475975382085110035}{562949953421312v^{17}}+\frac{53392138323683746235}{4503599627370496v^{18}}+\frac{26275374869163335461975}{18014398509481984v^{19}}\\ &\textstyle-\frac{105772979046693606062363}{288230376151711744v^{20}}-\frac{64759060397977041632277937}{1152921504606846976v^{21}}+\frac{130175508110590141461339987}{9223372036854775808v^{22}}\\ &\textstyle+O\!\left(\frac1{v^{23}}\right)\tag{4} \end{align} $$

Answer 2

Consider $$A=\frac{\Gamma\left(\frac{v+1}{2}\right)}{\Gamma\left(\frac{v}{2}\right)}$$ and take logarithms $$\log(A)=\log\left(\Gamma\left(\frac{v+1}{2}\right)\right)-\log\left(\Gamma\left(\frac{v}{2}\right)\right)$$ and consider Stirling expansion for large values of $x$ $$\log \left(\Gamma(x)\right)=x (\log (x)-1)+\frac{1}{2} \left(\log \left(\frac{1}{x}\right)+\log (2 \pi )\right)+\frac{1}{12 x}-\frac{1}{360 x^3}+\frac{1}{1260 x^5}+O\left(\frac{1}{x^{13/2}}\right)$$ So, applying this to the two different arguments, we have $$\log\left(\Gamma\left(\frac{v+1}{2}\right)\right)=\frac{1}{2} v (\log (v)-1-\log (2))+\frac{1}{2} \log (2 \pi )-\frac{1}{12 v}+\frac{7}{360 v^3}-\frac{31}{1260 v^5}+O\left(\frac{1}{v^{13/2}}\right)$$

$$\log\left(\Gamma\left(\frac{v}{2}\right)\right)=\frac{1}{2} v (\log (v)-1-\log (2))+\left(\frac{1}{2} \log \left(\frac{1}{v}\right)+\log \left(2 \sqrt{\pi }\right)\right)+\frac{1}{6 v}-\frac{1}{45 v^3}+\frac{8}{315 v^5}+O\left(\frac{1}{x^{13/2}}\right)$$ All of that leads to $$\log(A)=-\frac{1}{2} \log \left(\frac{2}{v}\right)-\frac{1}{4 v}+\frac{1}{24 v^3}-\frac{1}{20 v^5}+O\left(\frac{1}{v^{13/2}}\right)$$ Now, using $A=e^{\log(A)}$ and Taylor series again $$A=\frac{\sqrt{v}}{\sqrt{2}}-\frac{\left(\frac{1}{v}\right)^{1/2}}{4 \sqrt{2}}+\frac{\left(\frac{1}{v}\right)^{3/2}}{32 \sqrt{2}}+\frac{5 \left(\frac{1}{v}\right)^{5/2}}{128 \sqrt{2}}-\frac{21 \left(\frac{1}{v}\right)^{7/2}}{2048 \sqrt{2}}-\frac{399 \left(\frac{1}{v}\right)^{9/2}}{8192 \sqrt{2}}+\frac{869 \left(\frac{1}{v}\right)^{11/2}}{65536 \sqrt{2}}+O\left(\frac{1}{v^6}\right)$$

For the coefficients, you could find them in sequences $A143503$ and $A12854$ in $OEIS$.

Answer 3

Here is a method to obtain the full series of $\Gamma(a)/\Gamma(a+b)$, for large $a$, for any $b>0$ (one can then use recursion relations to access other values of $b$, for example). We have $$ \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = B(a,b) = \int_0^1 x^{a-1} (1-x)^{b-1} \, dx, $$ where $B$ is the beta function, and is defined by the right-hand side. Therefore if we can find the series for the right-hand side, we can just divide by $\Gamma(b)$ to get the series we want. Putting $x=e^{-y}$, this becomes $$ \frac{\Gamma(a)}{\Gamma(a+b)} = \frac{1}{\Gamma(b)}\int_0^{\infty} e^{-ay} (1-e^{-y})^{b-1} \, dx. $$ This integral is now in the form on which we can use Watson's lemma; the art is in obtaining the series for $(1-e^{-y})^{b-1}$. To find the first few terms, we can write $ 1-e^{-y}=y-y^2/2+\dotsb $ and use the binomial theorem, which tells us instantly that the leading order term is $$\frac{1}{\Gamma(b)}\int_0^{\infty} y^{b-1} e^{-ay} \, dy = a^{-b}. $$ I don't think there is an easy way to obtain the rest of the series for $(1-e^{-y})^{b-1}$ when $b$ is not an integer, but it will be of the form $ \sum_{n=0}^{\infty} c_n y^{b+n-1} $, with $c_0=1$, and then Watson's lemma gives $$ \frac{\Gamma(a)}{\Gamma(a+b)} \sim \sum_{n=0}^{\infty} c_n \frac{\Gamma(b+n)}{\Gamma(b)} a^{-b-n}. $$