A mathematical journey

It is given that:

$\begin{aligned} a_{n+1} & = 2na_n \\ \Rightarrow a_n & = 2(n-1)a_{n-1} \\ & = 2^2(n-1)(n-2) a_{n-2} \\ & = 2^3(n-1)(n-2)(n-3) a_{n-3} \\ & ... \quad ... \quad ... \\ & = 2^{n-1}(n-1)(n-2)(n-3)....(3)(2)(1) a_1 \quad \quad \color{#3D99F6}{\text{Since }a_1 = 1} \\ \Rightarrow a_n & = 2^{n-1}(n-1)! \end{aligned}$

Therefore,

$\begin{aligned} S(x) & = \sum_{n=1}^\infty \frac{x^n}{a_n} = \sum_{n=1}^\infty \frac{x^n}{2^{n-1}(n-1)!} = x \sum_{n=1}^\infty \frac{x^{n-1}}{2^{n-1}(n-1)!} = x \sum_{n=0}^\infty \frac{\left(\frac{x}{2}\right)^n}{n!} = xe^{\frac{x}{2}} \end{aligned}$

$\begin{aligned} \Rightarrow \int_0^2 S(x) \space dx & = \int_0^2 xe^{\frac{x}{2}} \space dx \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \color{#3D99F6}{\text{By integration by parts}} \\ & = \left[x^2e^{\frac{x}{2}} \right]_0^2 - \int_0^2 2e^{\frac{x}{2}} \space dx \\ & = 4e - \left[4e^{\frac{x}{2}} \right]_0^2 = 4e - 4e + 4 = \boxed{4} \end{aligned}$