Reckon you can solve this recursion?

$a_1 = 1 \quad a_2 = 2 \\ \Rightarrow a_3 = \dfrac {2(3^2)}{3-1} = \dfrac {3\times 3!}{2} \quad \Rightarrow a_4 = \dfrac {3\times 3!}{2} \times \dfrac{4^2}{3} = \dfrac {4\times 4!}{2} \\ \Rightarrow a_5 = \dfrac {4\times 4!}{2} \times \dfrac{5^2}{4} = \dfrac {5\times 5!}{2} \quad \Rightarrow a_n = \dfrac {n\times n!}{2} \\ \begin{aligned} \Rightarrow s & = \sum_{n=1}^\infty {\frac{n}{a_n}} = \frac{1}{1} + \frac{2}{2} + \sum_{n=3}^\infty {\frac{2n}{n\times n!}} = 2 + \sum_{n=3}^\infty {\frac{2}{n!}} \\ & = 2 + 2\sum_{n=0}^\infty {\frac{1}{n!}} - \frac{2}{0!} - \frac{2}{1!} - \frac{2}{2!} \\ & = 2 + 2e - 2 - 2 - 1 = 2e-3 = 2.436563657 \end{aligned}$

$\Rightarrow \lfloor 10^8\times s \rfloor = 243656\boxed{365}$

Define $b_{n}=\frac{a_{n}}{n}$ .

The recurrence then becomes:

$b_{n}=n*b_{n-1}$

This is obviously the recurrence relation for the factorial function.

Thus, we have,

$b_{n}=c*n! \quad \rightarrow a_{n}=c(n*n!)$

where c is an arbitrary constant to be determined from the initial conditions.

This c comes out to be $\frac{1}{2}$ and thus, $a_{n}=\frac{1}{2}(n*n!)$

The required summation is then trivial and evaluates to $2(e-1)$ .

The required answer is then $\boxed{365}$ .