<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] Well, you can carefully multiply and don't forget any terms:
$$\left( 1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots \right)^2 = \left( \color{blue}{1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots }\right)\left(\color{red}{ 1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots }\right)$$
Expanding and combining terms of the same degree:


*

*constant term: $$\color{blue}{1} \cdot \color{red}{1} = 1$$

*linear term: $$\color{blue}{1} \cdot \color{red}{\tfrac{1}{3}x}+\color{blue}{\tfrac{1}{3}x} \cdot \color{red}{1} = \tfrac{2}{3}x$$

*quadratic term: $$\color{blue}{1} \cdot \color{red}{\tfrac{1}{24}x^2}+\color{blue}{\tfrac{1}{3}x} \cdot \color{red}{\tfrac{1}{3}x} + \color{blue}{\tfrac{1}{24}x^2} \cdot \color{red}{1} = \tfrac{7}{36}x^2$$

*...

</s>