<s>[INST] Can $a(n) = \frac{n}{n+1}$ be written recursively? Take the sequence $$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \dots$$
Algebraically it can be written as $$a(n) = \frac{n}{n + 1}$$
Can you write this as a recursive function as well?
A pattern I have noticed:


*

*Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.


I am currently in Algebra II Honors and learning sequences 
 [/INST] \begin{align*}
a_{n+1} &= \frac{n+1}{n+2}  \\
    &= \frac{n+2-1}{n+2}  \\
    &= 1 - \frac{1}{n+2}  \text{, so }  \\
1 - a_{n+1} &= \frac{1}{n+2}  \text{, }  \\
\frac{1}{1 - a_{n+1}} &= n+2  &[\text{and so } \frac{1}{1 - a_n} = n+1]\\
    &= n+1+1  \\
    &= \frac{1}{1- a_n} +1  \\
    &= \frac{1}{1- a_n} + \frac{1-a_n}{1-a_n}  \\
    &= \frac{2-a_n}{1- a_n}  \text{, then }  \\
1 - a_{n+1} &= \frac{1-a_n}{2- a_n}  \text{, and finally }  \\
a_{n+1} &= 1 - \frac{1-a_n}{2- a_n}  \\
    &= \frac{2-a_n}{2- a_n} - \frac{1-a_n}{2- a_n}  \\
    &= \frac{1}{2- a_n}  \text{.}
\end{align*}
</s>