<s>[INST] 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
 [/INST] Hint:
\begin{align}
\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot 9}{2\cdot 4 \cdot 6 \cdot8\cdot10}.\frac{x^6}{12} &=\frac{(1\cdot 3 \cdot 5 \cdot 7 \cdot 9)(2\cdot 4 \cdot 6 \cdot8\cdot10)}{(2\cdot 4 \cdot 6 \cdot8\cdot10)^2}.\frac{x^6}{12} \\
&=\frac{10!}{2^{10}(1\cdot 2 \cdot 3 \cdot4\cdot 5)^2} \frac{x^6}{12}\\
&=\frac{1}{2^{10}} \cdot \binom{10}{5} \frac{x^6}{2(6)}\\
&= \frac{1}{2^{2\cdot (6-1)}} \cdot \binom{2\cdot(6-1)}{6-1} \frac{x^6}{2(6)}
\end{align}
Edit:
$$ \frac{1}{2^{2(n-1)}}\frac{(2n-2)!}{((n-1)!)^2}\frac{x^{2n}}{2n}$$
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