Stuck at Evaluating the Riemann-Stieltjes Integral Let $$\alpha(x) = \left\lbrace \begin{array}{cc} 0 & x=0\\ \dfrac{1}{2^n} & \dfrac{1}{3^n} < x \leq \dfrac{1}{3^{n-1}}\quad n=1,2,...\end{array}\right.$$
Evaluate $$\int_{0}^{1}{x\mathrm{d}\alpha(x)}$$
Attempt: I dont know how to put this formally, but I know it is similar to problem of evaluating the R-S integral where $\alpha(x)$ is the jump function. 
So If I evaluate the function $\alpha(x)$ for some values of $n$ then the function looks like this $$\alpha(x) = \left\lbrace \begin{array}{cc} 0 & x=0\\ \vdots & \vdots \\\dfrac{1}{8} & \dfrac{1}{27} < x \leq \dfrac{1}{9} \\ \dfrac{1}{4} & \dfrac{1}{9} < x \leq \dfrac{1}{3} \\ \dfrac{1}{2} & \dfrac{1}{3} < x \leq 1 \end{array}\right.$$ So $f(x)=x$ is continuous and $\alpha(x)$ is monotonically increasing. Hence the answer should be $$\int_{0}^{1}{x \mathrm{d}\alpha(x)} = \sum_{n=1}^{\infty}{\dfrac{1}{3^n}\dfrac{1}{2^{n+1}}}=\dfrac{1}{2}\sum_{n=1}^{\infty}{\left(\dfrac{1}{6}\right)^n}=\dfrac{1}{10}$$
So my questions are:
(a) Is the answer correct? If yes, then how can I write this formally? 
(b) If the answer is wrong, what direction can you provide me to get the right answer?
Thanks!

I think your answer is quite right.  I'm going to outline an alternative, wimpy approach using integration by parts:
$$\int_0^1 x \, d\alpha(x) = [x \, \alpha(x)]_0^1 - \int_0^1 dx \, \alpha(x)$$
The integrated term is $1/2$, and the resulting integral is straightforward:
$$\int_0^1 x \, d\alpha(x) = \frac12 - \sum_{n=1}^{\infty} \frac{1}{2^n} \left ( \frac{1}{3^{n-1}}-\frac{1}{3^n}\right) = \frac12 - 2 \sum_{n=1}^{\infty} \frac{1}{6^n} = \frac{1}{10}$$
