I have always wondered whether the Taylor Series of Gamma Function exists or not. I tried to find it, but in vain. I googled for it, but couldn't find it. Has anyone ever found its Taylor Series?
[Math] Taylor Series of Gamma Function
calculusgamma functionsoft-questiontaylor expansion
Related Solutions
As have already been said there is no golden trick here. However it is sometimes useful to know some ways to manipulate series and functions.
Suppose we know $$\frac{1}{1-x}=\sum_{k=0}^\infty x^k,\qquad|x|<1\tag{1}$$ and $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}, \qquad\text{for all $x$}\tag{2}$$
Formally if we integrate (1) we get $$-\log(1-x)=\sum_{k=0}^\infty \frac{x^{k+1}}{k+1}=\sum_{k=1}^\infty \frac{x^{k}}{k},\qquad|x|<1$$ and then $$\log(1+x)=-\sum_{k=1}^\infty \frac{(-1)^kx^{k}}{k},\qquad|x|<1$$ Computing $\arctan x$ is similar since by (1) $$(\arctan x)'=\frac{1}{1+x^2}=\sum_{k\geq0} (-1)^kx^{2k}$$
Also if we use $\sin x= (e^{ix}-e^{-ix})/2i$ in conjunction to (2) we get $$2i\sin x = \sum_{k=0}^\infty \frac{(ix)^k}{k!}- \sum_{k=0}^\infty \frac{(-ix)^k}{k!} = \sum_{k=0}^\infty \frac{i^k(1-(-1)^k)x^k}{k!}$$ now $1-(-1)^{k} = 0 $ for even $k$, and $1-(-1)^{k} = 2$ otherwise. So $$2i\sin x = 2\sum_{k=0}^\infty \frac{i^{2k+1}x^{2k+1}}{(2k+1)!} = 2i\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$$ where in the last step we used $i^{2k+1}=i\cdot i^{2k}=i\cdot (-1)^k$, and we reach the expansion $$\sin x=\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{(2k+1)!} $$ Cosine is similar to sine.
Note that these are formal derivations - proofs needs justifications of the steps done.
My professor used to say:
You might want to do calculus in $\Bbb{R}$, but the functions themselves naturally live in $\Bbb{C}$. Euler was the first to discover that if you don't look at what they do everywhere in the complex plane, you don't really understand their habits.
This is as subjective as it gets, but it has always helped my intuition. In particular, you might think that some function is doing nothing wrong, so it should be analytic. Well, if it does nothing wrong in $\Bbb{R}$, look at what it does in $\Bbb{C}$! If also in $\Bbb{C}$ it does nothing wrong, then it is analytic. If in $\Bbb{C}$ it makes some mess, then you have to be careful also in $\Bbb{R}$. To quote my professor again:
Even in $\Bbb{R}$, and in the most practical and applied problems, you can hear distant echos of the complex behavior of the functions. It's their nature, you can't change it.
Best Answer
If you want the Taylor series, you basically need the $n^{th}$ derivative of $\Gamma(x)$. These express in terms of the polygamma function. Considering
$$d_n=\frac{\left[\Gamma(x)\right]^{(n)}}{\Gamma(x)}$$ the first terms are $$d_1=\psi ^{(0)}(x)$$ $$d_2=\psi ^{(0)}(x)^2+\psi ^{(1)}(x)$$ $$d_3=\psi ^{(0)}(x)^3+3 \psi ^{(1)}(x) \psi ^{(0)}(x)+\psi ^{(2)}(x)$$ $$d_4=\psi ^{(0)}(x)^4+6 \psi ^{(1)}(x) \psi ^{(0)}(x)^2+4 \psi ^{(2)}(x) \psi ^{(0)}(x)+3 \psi ^{(1)}(x)^2+\psi ^{(3)}(x)$$ which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.