Series and probability The task at hand:
Suppose there are 3 players (A,B,C), who have 3 different dices with the following chances (per roll) to win:
Player A has a dice with a $\frac{1}{3}$ chance to win 
Player B has a dice with a $\frac{2}{5}$ chance to win 
Player C has a dice with a $\frac{4}{7}$ chance to win 
Player A starts rolling the dice, followed by Player B, then C and then starting with A all over again until one player wins.
Calculate the chance of player A to win. 
My question is wether the following approach is valid or has a logical error:
$\sum_{n=0}^\infty (\frac{2}{3}\frac{3}{5}\frac{3}{7})^n(\frac{1}{3}) = \frac{1}{3}\sum_{n=0}^\infty (\frac{2\cdot3\cdot3}{3\cdot5\cdot7})^n= \frac{1}{3}\sum_{n=0}^\infty (\frac{18}{105})^n= \frac{1}{3}\frac{1}{1-(\frac{18}{105})}= \frac{1}{3}\frac{1}{(\frac{87}{105})} =\frac{105}{261} \sim 0.4023$

Best, I think, to work "recursively".  
Define:  If $X,Y\in \{A,B,C\}$, not necessarily distinct,  then $P(X,Y)$ denotes the probability that $X$ will eventually win given that $Y$ is starting.  Thus the answer you seek is $P(A,A)$.
As is standard, we will also let $q_X=1-p_X$ for $X\in \{A,B,C\}$
Consider the first roll.  $A$ wins or she doesn't.  This implies that $$P(A,A)=p_A\times 1+q_A\times P(A,B)$$
Similarly:
$$P(A,B)=p_B\times 0+q_B\times P(A,C)\quad \&\quad P(A,C)=p_C\times 0+q_C\times P(A,A)$$
Easy to see that this implies that $$P(A,A)=\frac {p_A}{1-q_Aq_Bq_C}$$
Using your numbers we confirm that $P(A,A)=\fbox {0.402298851}$
