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$t_n$ 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.

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    I do not believe these are in geometric progression... – 2017-02-23
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    Using some reverse calculations, the limit seems to be $\frac{7}{12}$ – 2017-02-23
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    It is a GP, you just made a mistake in your answer as far as I can tell – 2017-02-23
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    @mrnovice It can't be GP, since the common ratios are not rational, but the partial sums are. – 2017-02-23
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    yes I see what you mean now - the answer I've given would only seem to hold if we were just given $t_{1}$ and $s_{2}$ – 2017-02-23

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$$ s_{\small1}=\frac{\color{Blue}{2}}{\color{red}{4}},\quad s_{\small2}=\frac{\color{Blue}{4}}{\color{red}{7}},\quad s_{\small3}=\frac{\color{Blue}{6}}{\color{red}{10}},\quad s_{\small4}=\frac{\color{Blue}{8}}{\color{red}{13}} $$ Thus, most probably: $$ s_{\small n}=\frac{\color{Blue}{2n}}{\color{red}{3n+1}} \quad\Rightarrow\quad \lim_{n\to\infty}s_{\small n}=\frac{\color{Blue}{2}}{\color{red}{3}} $$

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    More elegant than my work. – 2017-02-23
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    @DougM: Thanks. and (+1) for U. – 2017-02-23
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    How did you get the formula for $s_n$ – 2017-02-23
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    @lakada: By calculating the two arithmetic sequences of the nominators and the denominators. – 2017-02-24
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    Is there a way to express it as $s_n = \frac{a(1-r^n)}{1-r}$? I don't understand what r would be as it doesn't seem constant. – 2017-02-24
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    The point is it's not a gp – 2017-02-24
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    @lakada: Unfortunately, cannot. Because if $s_{\small n}$ equals to what you wrote, then $t_{\small n}$ should be a geometric sequence, which is not. – 2017-02-24
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$s_2 = t_1 + t_2, s_4 = s_2 + t_3 + t_4$ 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 $s = \frac 23$