<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] Notice
$$
\begin{align}
\frac{1}{2\cdot 4}x^2 &= \frac{2!}{(2^1 1!)^2(2 \cdot 2)} x^2
= \frac12 \binom{2}{1} \int_0^x \left(\frac{t}{4}\right)^1 dt
\\
\frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8}x^4
&= \frac{6!}{(2^3 3!)^2 (2\cdot 4)} x^4
= \frac12 \binom{6}{3} \int_0^x \left(\frac{t}{4}\right)^3 dt\\
\frac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12}x^6
&= \frac{10!}{(2^5 5!)^2 (2\cdot 6)} x^6
= \frac12 \binom{10}{5} \int_0^x \left(\frac{t}{4}\right)^5 dt
\end{align}
$$
So aside from the first constant term $1$, the $k^{th}$ non-constant term has the form
$$a_k = \frac12\binom{2\ell}{\ell}\frac{x^{\ell+1}}{2^{2\ell+1}(\ell+1)} = \frac12\binom{2\ell}{\ell}\int_0^x \left(\frac{t}{4}\right)^\ell dt
$$ where $\ell = 2k-1$, the $k^{th}$ odd number.
If one compute the ratio of successive terms of $a_k$, one find
$$\frac{a_{k+1}}{a_k} = \frac{\binom{4k+2}{2k+1}(2k)}{\binom{4k-2}{2k-1}(2k+2)} \frac{x^4}{2^4}
= \frac{(4k+2)(4k+1)(4k)(4k-1)}{(2k+1)^2(2k)(2k+2)}\frac{x^4}{2^4}
= \frac{\left(1-\frac{1}{4k}\right)^2}{\left(1 + \frac{1}{2k}\right)\left(1+\frac1k\right)} x^4$$
Since this ratio converges to $x^4$ as $k \to \infty$, the radius of convergence of the series is $1$. 
In fact, for $|x| < 1$, the series has following representation
$$1 + \frac12\sum_{\ell\text{ odd}}\binom{2\ell}{\ell}\int_0^x \left(\frac{t}{4}\right)^\ell dt
= 1 + \frac14\sum_{\ell=0}^\infty \binom{2\ell}{\ell} (1 - (-1)^{\ell})\int_0^x \left(\frac{t}{4}\right)^\ell dt
$$
For $|z| < \frac14$, we have following identity:
$$\sum_{\ell=0}^\infty \binom{2\ell}{\ell} z^\ell = \frac{1}{\sqrt{1-4z}}$$
From this, we find for $|x| < 1$, the series evaluates to
$$1 + \frac14 \int_0^x \left(\frac{1}{\sqrt{1-t}} - \frac{1}{\sqrt{1+t}}\right) dt
= 2 - \frac12\left( \sqrt{1+x} + \sqrt{1-x}\right)$$
</s>