Earlier today, i had a simple test on linear equations. since i got done early, i started fooling around, trying to calculate the equation for the tables and charts. one equation that is proving to be really difficult to solve is the seemingly simple fraction series. it is $\frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \frac{1}{11},\cdots$
Anyway, i tried to figure this out, first by figuring out how to get from one fraction to another. it went like this:
$$ \frac{1}{5} \times \frac{5}{7} \Rightarrow \frac{1}{7} \times \frac{7}{9} \Rightarrow \frac{1}{9}\cdots $$
I later went through this and decided to look at the difference of each fraction $\left(\frac{5}{7} \Rightarrow \frac{7}{9} \right) \text{etc}$ to see if the pattern got to $0$. i soon realized that:
$$\begin{align} \frac{7}{9}-\frac{5}{7}&=\frac{4}{63}\\ \frac{9}{11}-\frac{7}{9}&=\frac{4}{99}\\ \frac{11}{13}-\frac{9}{11}&=\frac{4}{143}\\ \frac{15}{13}-\frac{13}{15}&=\frac{4}{195}\\ &\cdots \end{align}$$
the thing i noticed is that the numerator is always $4$. so i just needed to get the denominator to be $0$, and then the numerator should be the same, right?(not really) this is what i did: $$ \begin{align} 63 \quad 99 &\quad 143 \quad 195\\ 36 \quad &44 \quad 52\\ 8 &\quad 8 \\ &0 \end{align}$$
so now that I´ve proven there is a solution to this, i just need to know what the formula would be. can anyone pick up the slack and help? (i feel like I´m missing something) thanks in advance.
How to get from one fraction to another.
I set $a_n=\dfrac{1}{2n+1}=m$
solving for $n$ I got $n=\dfrac{1-m}{2 m}$
thus the next fraction that is $\dfrac{1}{2(n+1)+1}$ will be
$$\frac{1}{2 \left(\frac{1-m}{2 m}+1\right)+1}=\frac{m}{2 m+1}$$
therefore the recurrence is
$a_1=\dfrac{1}{5}$
$a_{n+1}=\dfrac{a_n}{2a_n+1}$
for instance substituting $a_n=1/13$ you get
$a_{n+1}=\dfrac{1/13}{2/13+1}=1/15$
Is this you were looking for?
Let me know...