Here is a tractable way forward to find the coefficients of the series for $\sec(x)$. It is straightforward to adopt this apply this approach to find the series for $9\sec(3x)$.
Recall that $\sec(x)\,\cos(x)=1$ and that the series for the cosine function is given by
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}$$
Furthermore note that since the secant function is even, its Taylor series will be given by
$$\sum_{n=0}^\infty \frac{a_nx^{2n}}{(2n)!}$$
Then, multiplying the series in $(1)$ with the series in $(2)$ yields
$$\begin{align}
1&=\sec(x)\,\cos(x)\\\\
&=\left(\sum_{m=0}^\infty \frac{a_mx^{2m}}{(2m)!}\right)\left(\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}\right)\\\\
&=\sum_{p=0}^\infty\left(\sum_{m=0}^p \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}\right)x^{2p}
\end{align}$$
Therefore, we must have
$$\sum_{m=0}^p \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}=\begin{cases}
1&,p=0\\\\
0&,p>0
\end{cases} \tag 1$$
We can use $(1)$ to find the coefficients $a_m$ recursively. We see that for $p=0$, $(1)$ reveals that $a_0=1$ and for $p>0$ we have the recursive relationship
$$\bbox[5px,border:2px solid #C0A000]{a_p=-(2p)!\sum_{m=0}^{p-1} \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}} \tag 2$$
Let's use $(2)$ to find the first few coefficients of the secant function. For $p=1$, we have
$$a_1=-(2)!\frac{(-1)^{1-0}a_0}{(0)!(2)!}=1$$
For $p=2$, we have
$$a_2=-(4)!\left(\frac{(-1)^{2-0}a_0}{(0)!((4)!)}+\frac{(-1)^{2-1}a_1}{(2)!(2)!}\right)=5$$
For $p=3$, we have
$$a_3=-(6)!\left(\frac{(-1)^{3-0}a_0}{(0)!((6)!)}+\frac{(-1)^{3-1}a_1}{(2)!(4)!}+\frac{(-1)^{3-2}a_2}{(4)!(2)!}\right)=61$$
We can continue recursively to obtain coefficients for higher order terms, but are content here to write the series using the firsts few terms as
$$\bbox[5px,border:2px solid #C0A000]{\sec(x)=1+\frac x2+\frac{5x^2}{24}+\frac{61x^4}{720}+R_8(x)}$$
where the remainder $R_8(x)$ is
$$R_8(x)=\sum_{p=4}^\infty \left(\sum_{m=0}^p \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}\right)x^{2p}=O(x^8)
$$
Best Answer
Use $\Gamma(z)=(z-1)\Gamma(z-1)$, and $\Gamma(1/2)=\sqrt\pi$, we get
$$\Gamma(n+\frac{1}{2})=(n-\frac{1}2)\cdots\frac{3}2\cdot\frac{1}2\cdot\sqrt\pi=\frac{(2n-1)!!}{2^n}\sqrt\pi=\frac{(2n)!}{(2n)!!\cdot2^n}\sqrt\pi$$
Note that $(2n)!!=2^n n!$, we get
$$\begin{align}\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}(2n+1)n!}x^{2n+1}&=\sum_{n=0}^{\infty} \frac{(2n)!}{2^n n!\cdot2^n}\sqrt\pi\cdot\frac{1}{\sqrt{\pi}(2n+1)n!}x^{2n+1}\\ \\ &=\sum_{n=0}^{\infty} \frac{(2n)!}{(2^n n!)^2 (2n+1)}x^{2n+1}\end{align}$$
So they are equivalent.