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$$\frac {2}{3}, \frac {14}{18}, \frac {43}{54},\frac {259}{324},\frac {1555}{1944},\frac {9331}{11664} \frac {55987}{69984}...$$ The pattern that I got for the numerators (not too clean) was, the first term, multiply by 7, then the next term multiply by 3 and add 1. Then from there it's all multiply by 6 and 1. The denominators are clearly just multiplying by 6.

But how can I find the limit sequence? Or I guess more simply put, how would you generate the recursive formula for this?

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  • $\begingroup$ The denominators are not just multiplying by $6$. $54 = 3 \times 18$. $\endgroup$ – JimmyK4542 Jan 3 '17 at 9:17
  • $\begingroup$ Yes, you're right, i'm sorry, I forgot to mention that. Are there other ways to form these sequences? It doesn't seem right to find a pattern that holds most of the time.. $\endgroup$ – Ryan Goulden Jan 3 '17 at 9:18
  • $\begingroup$ Try multiplying the $n$-th term by $6^n$ and see if you can find a pattern. Also, where are you getting these fractions from? $\endgroup$ – JimmyK4542 Jan 3 '17 at 9:19
  • $\begingroup$ The terms come from the n-th composition of the expression of the form $$\frac {1+\frac {1+\frac {1+\frac {1+...}{3}}{2}}{3}}{2}$$ Where I'm nesting $$\frac{1+\frac{1}{3}}{2}$$ $\endgroup$ – Ryan Goulden Jan 3 '17 at 9:24
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    $\begingroup$ The $n$-th term appears to be $\dfrac{4}{5}(1-6^{-n})$. $\endgroup$ – JimmyK4542 Jan 3 '17 at 9:26
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Since you are nesting $\dfrac{1+\dfrac{1+\cdots}{3}}{2}$, if we let $x_n$ be the $n$-th term in your series, then we have \begin{align}x_{n+1} &= \dfrac{1+\dfrac{1+x_n}{3}}{2} \\ x_{n+1} &= \dfrac{1}{6}x_n+\dfrac{2}{3}\end{align}

This is a non-homogeneous linear recurrence relation. We also have the initial condition $x_1 = \dfrac{2}{3}$.

I'm not sure if you are familiar with methods for solving linear recurrence relations, but even if you aren't, you can show that $$x_n = \dfrac{4}{5}(1-6^{-n})$$ by using induction.

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The pattern is:

$$\frac {2}{3},\frac {2*6+2}{3*6}=\frac {14}{18},\frac {14*6+2}{18*6}= \frac {86}{108},\frac {86*6+2}{108*6}=\frac {518}{648},\frac {518*6+2}{648*6}=\frac {3110}{3888},...$$

Let's work better each term:

$$\frac {2}{3}$$

$$\frac {2*6+2}{3*6}=\frac {2}{3}+\frac {2}{3*6}=\frac {14}{18}$$

$$\frac {14*6+2}{18*6}=\frac {2}{3}+\frac {2*6+2}{3*6*6}=\frac {86}{108}$$

$$......$$

Now, applying geometric progression, the sequence tends to:

$$\frac {2}{3}+\frac {2*6*6+2*6+2...}{3*6*6*6...}= \frac {2}{3}+\frac{2}{3} * \frac{1}{5}=0.8$$

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