Geometric series and sequences $t_n$ is the $n^{th}$ term of an infinite sequence and $s_n$ is the partial sum of the first $n$ terms. Given that $t_1 = \frac{2}{4}$, $s_2 = \frac{4}{7}$, $t_3 = \frac{1}{35}$, $s_4 = \frac{8}{13}$. Determine a formula for $s_n$ and calculate its limit.
 A: $$ s_{\small1}=\frac{\color{Blue}{2}}{\color{red}{4}},\quad s_{\small2}=\frac{\color{Blue}{4}}{\color{red}{7}},\quad s_{\small3}=\frac{\color{Blue}{6}}{\color{red}{10}},\quad s_{\small4}=\frac{\color{Blue}{8}}{\color{red}{13}} $$ 
Thus, most probably: 
$$ s_{\small n}=\frac{\color{Blue}{2n}}{\color{red}{3n+1}} \quad\Rightarrow\quad \lim_{n\to\infty}s_{\small n}=\frac{\color{Blue}{2}}{\color{red}{3}} $$ 
A: $s_2 = t_1 + t_2, s_4 = s_2 + t_3 + t_4$ this is enough information to find $t_2, t_4$
$t_1,t_2,t_3, t_4 =$$ \frac 12, \frac 1{14}, \frac 1{35}, \frac 1{65}\\
\frac {1}{2\cdot 1}, \frac {1}{2\cdot 7}, \frac {1}{5\cdot 7},\frac {1}{5\cdot 13}$
Just a hunch, but it appears that the first factor in the denominator increases by $3$ every other term, and the second factor increases by $6$ every second term.
$t_{2k-1} = \frac {1}{(3k-1)(6k-5)}\\  
t_{2k} = \frac {1}{(3k-1)(6k+1)}$
$s_{2k} = $$s_{2k-2} + \frac {1}{(3k-1)(6k-5)} + \frac {1}{(3k-1)(6k+1)}\\
s_{2k-2} + \frac {12k-4}{(3k-1)(6k-5)(6k+1)}\\ s_{2k-2} + \frac {4}{(6k-5)(6k+1)}\\ s_{k-2} + \frac {2}{3(6k-5)} - \frac {2}{3(6k+1)}\\
\sum_\limits{i=1}^{k} \left(\frac {2}{3(6i-5)} - \frac {2}{3(6i+1)}\right)$
We have a telescoping series.
$s_{2k} = \frac 23 - \frac {2}{3(6k+1)}$
as k goes to infinity
$s = \frac 23$
