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.

So it doesn't explicitly use generating functions as I supposed, but it is pretty close. 
You have (for ease of visualization):
$$\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)^2=\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)=\sum_{n=0}^{\infty}\left[\frac{2}{n!(n+2)!}x^n\left(\sum_{j=0}^{\infty}\frac{2}{j!(j+2)!}x^j\right)\right].$$
I think it is fairly obvious that each power of $x$ will appear, since all coefficients are positive and such. But we will soon see that this is the case anyhow.
To begin writing out the result explicitly, we need to consider each power of $x$. To get the constant term (ie, the $x^0$ term), the only option is $n=j=0$ and so we just have $\frac{2}{0!(0+2)!}\cdot\frac{2}{0!(0+2)!}=1$, so our first term in the result is $1$.
Next, we need the $x^1$ term, and there are two ways to do this. We get $x^1$ when $n=0, j=1$ and when $n=1, j=0$. So adding the two results gives $\frac{2}{0!(0+2)!}\cdot\frac{2}{1!(1+2)!}+\frac{2}{1!(1+2)!}\cdot\frac{2}{0!(0+2)!}=\frac{2}{3},$ so the next term in our result is $\frac{2}{3}x$.
I'll do one more. For the $x^2$ term, we have three options: $n=j=1$, or $n=2, j=0$, or $n=0, j=2$. I hope by now it is clear that if $j\neq n$, we can just double the result, so our coefficient is: $\frac{2}{1!(1+2)!}\cdot\frac{2}{1!(1+2)!}+(2)\cdot\frac{2}{2!(2+2)!}=\frac{3}{37}$.
So, to find the coefficient for a general $x^k$, you need to find all the ways $n+j=k$ in non-negative integers $n,j$, and then add all such results up. It gets worse as you go. There are ${k+1 \choose k-1}$ ways to do this for a given $k>0$. If $k=10$, we have $55$ ways to consider!
But now you can give the result clearly to whatever accuracy you'd like. And with a little effort, you could find a closed form for the result, too.
