Proving an irrational number in the cantor set I'm trying to prove that $0.2020020002\ldots_3 \in \Bbb Q^c\cap C$ where $C$ denotes the Cantor set. I'm trying to get a contradiction assuming $0.2020020002\ldots_3 \in \Bbb Q$ (without using the fact that every rational number is periodic or terminating).
Suppose $$\sum_{k=1}^{\infty} \frac{2}{3^{\frac{k(k+1)}{2}}}=\frac{p}{q}.$$  
Choose $n$ such that $3^n \gt q+1$. Then $$ q \times3^n\times \sum_{k=1}^{\infty} \frac{2}{3^{\frac{k(k+1)}{2}}}= p \times 3^n=integer.$$ I'm stuck after this step. How can I show that this is an integer between $0$ and $1$?
Any help is greatly appreciated.

Suppose $$ \sum_{k=1}^{\infty}\frac{2}{3^{\frac{k(k+1)}{2}}} \text{ is rational.}$$ Then there exists $ p,q\in \mathbb{Z}^{+} $ such that  $\gcd(p,q)=1$ and
$$ \sum_{k=1}^{\infty}\frac{2}{3^{\frac{k(k+1)}{2}}}=\frac{p}{q} .$$
Then you need to choose $ n\in \mathbb{N} $ such that $ n=\frac{k_{0}(k_{0}+1)}{2} $ for some $ k_{0}\in \mathbb{N} $ with $ 3^{k_{0}}>q+1 $.
Put $$ x=q3^{n}\left( \frac{p}{q}-\sum_{k=1}^{k_{0}}\frac{2}{3^{\frac{k(k+1)}{2}}}\right). $$
Then $$x=3^{n}p-q3^{n}\sum_{k=1}^{k_{0}}\frac{2}{3^{\frac{k(k+1)}{2}}}. $$
Hence $x$ is an integer.
Observe that $$x=q3^{n}\sum_{k=k_{0}+1}^{\infty}\frac{2}{3^{\frac{k(k+1)}{2}}}>0. $$
Therefore $x$ is a positive integer.
Also observe that $$x=q3^{n}\sum_{k=k_{0}+1}^{\infty}\frac{2}{3^{\frac{k(k+1)}{2}}}<q3^{n}\sum_{k=k_{0}+1}^{\infty}\frac{2}{3^{n+k}}<q\sum_{k=k_{0}+1}^{\infty}\frac{2}{3^{k}}=\frac{q}{3^{k_{0}}}<\frac{q+1}{3^{k_{0}}}<1. $$
That is $x$ is a positive integer with $ x<1 $. So we have a contradiction.
