Approximation of the Gamma function for small value

Using Taylor series around $u=0$, you should get $$\Gamma(u)=\frac{1}{u}-\gamma +\frac{6 \gamma ^2+\pi ^2}{12} u+O\left(u^2\right)\tag 1$$

For $u=\frac 1 {10}$, this very limited expression would give $\approx 9.52169$ while the "exact" value would be $\approx 9.51351$.

The error is lower than $0.1$% for any $0\lt x\leq \frac 1 {10}$.

Edit

It is possible to slightly improve the above approximation builging the simplest Pade approximant of $u\, \Gamma(u)$ around $u=0$. This would lead to $$\Gamma(u)=\frac 1 u \times\frac{12\gamma+(\pi^2-6\gamma^2)u}{12\gamma+(\pi^2+6\gamma^2)u}\tag 2$$ Using $(1)$ leads to overestimates while using $(2)$ leads to underestimates which makes the average much better. So, a better approximation could be $$\Gamma(u)=\frac 1 {2u}\left(1-\gamma u+\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) u^2+\frac{12\gamma+(\pi^2-6\gamma^2)u}{12\gamma+(\pi^2+6\gamma^2)u} \right)\tag 3$$ for which the error is lower than $0.01$% for any $0\lt x\leq \frac 1 {10}$.

We could also consider $$\Gamma(u)=\alpha\left(\frac{1}{u}-\gamma +\frac{6 \gamma ^2+\pi ^2}{12} u \right)+(1-\alpha)\left(\frac 1 u \times\frac{12\gamma+(\pi^2-6\gamma^2)u}{12\gamma+(\pi^2+6\gamma^2)u}\right)$$ and optimize the $\alpha$ parameter. For the consider range $\alpha\approx 0.44$ seems to be quite good leading to a maximum error lower than $0.0002$% over that range.

If we consider $0< x \leq 1$, $\alpha\approx 0.35$ leads to errors smaller than $0.6$% for the entire range.

We could also show that the Padé approximant of $u\,\Gamma(u)$ lead to relative errors lower than $1$% for the range $-0.625 \leq x \leq 1.168 $.

$\Gamma(z)$ has a pole at zero, with Laurent series $$ \frac{1}{z}-\gamma+ \left( {\frac {{\pi }^{2}}{12}}+{\frac {{\gamma}^{2} }{2}} \right) z+ \left( -{\frac {\zeta \left( 3 \right) }{3}}-{\frac {{\pi }^{2}\gamma}{12}}-{\frac {{\gamma}^{3}}{6}} \right) {z}^{2}+ \left( {\frac {{\pi }^{4}}{160}}+{\frac {\zeta \left( 3 \right) \gamma}{3}}+{\frac {{\pi }^{2}{\gamma}^{2}}{24}}+{\frac {{\gamma}^{4} }{24}} \right) {z}^{3}+ \left( -{\frac {\zeta \left( 5 \right) }{5}}- {\frac {{\pi }^{4}\gamma}{160}}-{\frac {\zeta \left( 3 \right) {\pi } ^{2}}{36}}-{\frac {\zeta \left( 3 \right) {\gamma}^{2}}{6}}-{\frac {{ \pi }^{2}{\gamma}^{3}}{72}}-{\frac {{\gamma}^{5}}{120}} \right) {z}^{4 }+ \left( {\frac {61\,{\pi }^{6}}{120960}}+{\frac {\zeta \left( 5 \right) \gamma}{5}}+{\frac {{\pi }^{4}{\gamma}^{2}}{320}}+{\frac { \left( \zeta \left( 3 \right) \right) ^{2}}{18}}+{\frac {\zeta \left( 3 \right) {\pi }^{2}\gamma}{36}}+{\frac {\zeta \left( 3 \right) {\gamma}^{3}}{18}}+{\frac {{\pi }^{2}{\gamma}^{4}}{288}}+{ \frac {{\gamma}^{6}}{720}} \right) {z}^{5}+O \left( {z}^{6} \right) $$ Of course $\gamma$ is Euler's constant and $\zeta$ is Riemann's zeta function.

I wonder if it looks better if those $\pi^2$ and $\pi^4$ terms are written in terms of $\zeta(2)$ and $\zeta(4)$ and so on?

$$ \frac{1}{z}-\gamma+ \left( {\frac {\zeta \left( 2 \right) }{2}}+{\frac { {\gamma}^{2}}{2}} \right) z+ \left( -{\frac {\zeta \left( 3 \right) }{3}}-{\frac {\zeta \left( 2 \right) \gamma}{2}}-{\frac {{\gamma}^{3} }{6}} \right) {z}^{2}+ \left( {\frac {9\,\zeta \left( 4 \right) }{16}} +{\frac {\zeta \left( 3 \right) \gamma}{3}}+{\frac {\zeta \left( 2 \right) {\gamma}^{2}}{4}}+{\frac {{\gamma}^{4}}{24}} \right) {z}^{3}+ \left( -{\frac {\zeta \left( 5 \right) }{5}}-{\frac {9\,\zeta \left( 4 \right) \gamma}{16}}-{\frac {\zeta \left( 3 \right) \zeta \left( 2 \right) }{6}}-{\frac {\zeta \left( 3 \right) {\gamma}^{2}}{ 6}}-{\frac {\zeta \left( 2 \right) {\gamma}^{3}}{12}}-{\frac {{\gamma} ^{5}}{120}} \right) {z}^{4}+ \left( {\frac {61\,\zeta \left( 6 \right) }{128}}+{\frac {\zeta \left( 5 \right) \gamma}{5}}+{\frac {9 \,\zeta \left( 4 \right) {\gamma}^{2}}{32}}+{\frac { \left( \zeta \left( 3 \right) \right) ^{2}}{18}}+{\frac {\zeta \left( 3 \right) \zeta \left( 2 \right) \gamma}{6}}+{\frac {\zeta \left( 3 \right) { \gamma}^{3}}{18}}+{\frac {\zeta \left( 2 \right) {\gamma}^{4}}{48}}+{ \frac {{\gamma}^{6}}{720}} \right) {z}^{5}+O \left( {z}^{6} \right) $$

Maybe part of the $\zeta(4)$ should be a $\zeta(2)^2$ and similarly for the $\zeta(6)$ to make the pattern recognizable?