Sequence using BT

By binomial theorem , we have

$\begin{aligned} (1+x)^{20} & = x^{20} + \frac {20!}{1!19!}x^{19} + \frac {20!}{2!18!}x^{18} + \cdots + \frac {20!}{18!2!}x^2 + \frac {20!}{19!1!}x + 1 \\ (1-x)^{20} & = x^{20} - \frac {20!}{1!19!}x^{19} + \frac {20!}{2!18!}x^{18} - \cdots + \frac {20!}{18!2!}x^2 - \frac {20!}{19!1!}x + 1 \\ (1+x)^{20} - (1-x)^{20} & = 2\left(\frac {20!}{1!19!}x^{19} + \frac {20!}{3!17!}x^{17} + \cdots + \frac {20!}{17!2!}x^3 + \frac {20!}{19!1!}x \right) \end{aligned}$

Now putting $x=1$ and divide both sides by $2\cdot 20!$ :

$\frac 1{1!19!} + \frac 1{3!17!} + \frac 1{5!15!} + \cdots + \frac 1{19!1!} = \boxed{\frac {2^{19}}{20!}}$

If we take the expression given and multiply by $20!$ , we get $\sum_{k=1}^{10} \binom{20}{2k-1}$ ,

i.e. you are getting the total number of subsets of a set with twenty elements that have an odd number of elements.

Now $2^{20}$ is the total number of subsets of a set with $20$ elements, and exactly half of them have an odd number of elements. (An obvious fact for $n$ odd, and a proof by induction is pretty straightforward.)

So $2^{19}$ is the total number of subsets of a set with $20$ that themselves have an odd number of elements. I.e.,

$2^{19} = \sum_{k=1}^{10} \binom{20}{2k-1}$ . Divide by $20!$ to get the identity above.

$\begin{aligned} S &= \dfrac{1}{20!} \{\dfrac{20!}{1!19!} + \dfrac{20!}{3!17!} + \dfrac{20!}{5!15!} + \dots + \dfrac{20!}{19!1!}\} \\ &= \dfrac{1}{20!} \{\dbinom{20}{1} + \dbinom{20}{3} + \dbinom{20}{5} + \dots + \dbinom{20}{19}\} \\ &= \dfrac{1}{20!} (2^{20 - 1}) \\ &= \boxed{\dfrac{2^{19}}{20!}} \end{aligned}$