Convergence and divergence of an infinite series

The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do different series test but here I am unable to find the nth term of the given series please help me out of this.
The question is different as it contains x terms and its nth term will be totally different from marked as duplicate question

Answer

Hint 1: Since $(4n-2)^2\gt(4n-1)(4n-3)$, we have
$$\begin{align} \frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)} &\le\sqrt{\frac{1\cdot3\cdots(4n-3)}{3\cdot5\cdots(4n-1)}\cdot\frac{1\cdot3\cdots(4n-3)}{1\cdot3\cdots(4n-3)}}\\ &=\sqrt{\frac1{4n-1}} \end{align}$$
Hint 2: Since $(4n-3)^2\gt(4n-2)(4n-4)$, we have
$$\begin{align} \frac12\cdot\frac{3\cdot5\cdots(4n-3)}{4\cdot6\cdots(4n-2)} &\ge\frac12\sqrt{\frac{4\cdot6\cdots(4n-2)}{4\cdot6\cdots(4n-2)}\cdot\frac{2\cdot4\cdots(4n-4)}{4\cdot6\cdots(4n-2)}}\\ &=\frac12\sqrt{\frac1{2n-1}} \end{align}$$

---

Thus,
$$\frac12\sqrt{\frac1{2n-1}}\,\frac{x^{2n}}{4n}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\sqrt{\frac1{4n-1}}\,\frac{x^{2n}}{4n}$$
and therefore, since $4n-1\ge3n$ for $n\ge1$, we have

$$\bbox[5px,border:2px solid #C0A000]{\frac{x^{2n}}{8\sqrt2\,n^{3/2}}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\frac{x^{2n}}{4\sqrt3\,n^{3/2}}}$$