<s>[INST] How to show $\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$? I tried to find the right handside of the equation by manipulating the series but I failed at getting the right handside of it.
$$\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$$
Closed form of the left handside in the parantheses would be
$$\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n$$
Any hint would be appreciated.
 [/INST] We first calculate the Cauchy product/product of power series
\begin{align*}
\left(\sum_{n=0}^\infty \frac{2}{n!(n+2)!}\cdot x^n\right)^2 &= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{2}{k!(k+2)!}\cdot \frac{2}{(n-k)!(n-k+2)!} \right)\cdot x^n\\
&= \sum_{n=0}^\infty \left(\frac{4}{(n+2)!^2}\cdot \sum_{k=0}^n \binom {n+2}{k} \cdot \binom{n+2}{k+2}\right)\cdot x^n.
\end{align*}
By the Vandermonde convolution identity we have
$$
\sum_{k=0}^n\binom{n+2}{k}\cdot\binom{n+2}{k+2} = \sum_{k=0}^n \binom{n+2}{k} \cdot \binom{n+2}{n-k} = \binom{2n+4}{n}.
$$
Thus, we have
$$
\left(\sum_{n=0}^\infty \frac{2}{n!(n+2)!}\cdot x^n\right)^2 = \sum_{n=0}^\infty \frac{4}{(n+2)!^2}\cdot \binom{2n+4}{n}\cdot x^n.
$$
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