With the help of Wolfram Alpha I got the following $$\frac{\sum_{i=x+1}^{\infty} i q^i}{\sum_{i=1}^x i q^i}=\frac{q^x (x(1-q))+1}{q^x (x(q-1)-1)+1}$$
Now I am intersted in the behavior of this sum for $0<q<1$ a constant and $x \rightarrow \infty$.
Here, Wolfram Alpha does not help.
By dividing by $q^x$ I get: $$\frac{x(1-q)+q^{-x}}{x(q-1)-1+q^{-x}}\rightarrow 1$$ Is this true?
Note that $\sum_{i=1}^x iq^i \to \sum_{i=1}^\infty iq^i = q/(1-q)^2$, and hence, as this sum converges, $\sum_{i=x+1}^\infty iq^i \to 0$, $x\to\infty$.
That gives $$ \frac{\sum_{i=x+1}^\infty iq^i}{\sum_{i=1}^x iq^i} \to 0 $$ Note that Wolfram|Alpha gives you $$ \frac{\sum_{i=x+1}^\infty iq^i}{\sum_{i=1}^x iq^i} = \frac{q^x(-qx+x+1)}{xq^{x+1} -(x+1)q^x + 1} \to 0 $$ (you missed one paranthesis).