$\frac{2^2}{3.4}x^4 + \frac{2^2.4^2}{3.4.5.6}x^6+\cdots = \sum a_n$ Given the following series: 
$$\frac{2^2}{3.4}x^4 + \frac{2^2.4^2}{3.4.5.6}x^6+\cdots = \sum a_n,$$ where $$a_n = \frac{2^2.4^2\cdots(2n)^2}{3.4.5.6\cdots (2n+2)} x^{2n+2}$$
Now we try Ratio test first: $$\frac{a_{n+1}}{a_n} = \frac{2n^2+4n+2}{n+2}x^2$$
but then I cant conclude from that, how to proceed next?
 A: Apply Ratio Test
$$
\begin{align}
\frac{2^2}{3\cdot4}x^4 + \frac{2^2\cdot4^2}{3\cdot4\cdot5\cdot6}x^6+\cdots
&=\sum_{k=2}^\infty\frac{(k-1)!^22^{2k-2}}{(2k)!/2!}x^{2k}\\
&=\frac12\sum_{k=2}^\infty\frac{(2x)^{2k}}{k^2\binom{2k}{k}}\\
\end{align}
$$
The Ratio Test says
$$
\begin{align}
\lim_{k\to\infty}\frac{a_{k+1}}{a_k}
&=\lim_{k\to\infty}\frac{k^2}{(k+1)^2}\frac{4x^2}{\frac{(2k+2)(2k+1)}{(k+1)^2}}\\
&=\lim_{k\to\infty}\frac{4k^2}{(2k+2)(2k+1)}x^2\\[9pt]
&=x^2
\end{align}
$$
Therefore, the series converges for $|x|\lt1$.

The Case $\boldsymbol{|x|=1}$
Using the estimate $(9)$ given in this answer, we have
$$
\begin{align}
\frac12\frac{4^k}{k^2\binom{2k}{k}}
&\le\frac12\frac{4^k}{k^2\frac{4^k}{\sqrt{\pi(k+1/3)}}}\\
&=\frac{\sqrt{\pi(k+1/3)}}{2k^2}\\[9pt]
&\le\frac{\sqrt{\pi/3}}{k^{3/2}}
\end{align}
$$
So the series converges for $|x|=1$.

Applying Raabe's Test
Another approach for the case $|x|=1$ is to use Raabe's Test:
$$
\begin{align}
\lim_{k\to\infty}k\left(\frac{a_k}{a_{k+1}}-1\right)
&=\lim_{k\to\infty}k\left(\frac{(2k+2)(2k+1)}{4k^2}-1\right)\\
&=\lim_{k\to\infty}k\left(\frac{6k+2}{4k^2}\right)\\[3pt]
&=\frac32
\end{align}
$$
which is greater than $1$, so the series converges for $|x|=1$.

The Actual Sum
Applying the substitution $x\mapsto4x^2$ to this answer gives
$$
\sum_{k=0}^\infty\frac{(2x)^{2k}}{\binom{2k}{k}}
=\frac1{1-x^2}\left[1+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right]
$$
Integrating once gives
$$
\frac12\sum_{k=0}^\infty\frac{(2x)^{2k+1}}{(2k+1)\binom{2k}{k}}
=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}
$$
Integrating again gives
$$
\frac14\sum_{k=0}^\infty\frac{(2x)^{2k+2}}{(2k+2)(2k+1)\binom{2k}{k}}
=\frac{\sin^{-1}(x)^2}{2}
$$
Therefore,
$$
\begin{align}
\frac12\sum_{k=2}^\infty\frac{(2x)^{2k}}{k^2\binom{2k}{k}}
&=\frac12\sum_{k=1}^\infty\frac{(2x)^{2k+2}}{(2k+2)(2k+1)\binom{2k}{k}}\\[6pt]
&=\sin^{-1}(x)^2-x^2
\end{align}
$$
