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 
 A: \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*}
A: After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = \frac{1}{2 - A_{n}}$$ where $$A_1 = \frac{1}{2}$$
A: Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = \frac{na_{n-1} + 1}{n+1}$$
with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).
A: Perhaps a bit simpler is to note that
$$
\overbrace{\ \frac1{1-a_n}\ }^{n+1}+1=\overbrace{\frac1{1-a_{n+1}}}^{n+2}\\
$$
solving for $a_{n+1}$ yields
$$
\frac1{2-a_n}=a_{n+1}\\
$$
A: You can, and in multiple ways. Such as:$$a_n=\frac{1}{a_{n-1}+\frac{2}{n}}\tag{1}$$
or 
$$a_n=\frac{na_{n-1}+1}{n+1}\tag{2}$$
etc.
equation 1, is simply noting: $$\frac{n-1}{n}+\frac{2}{n}=\frac{n+1}{n}=\frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$
and equation 2, simply notes:
$$n=na_{n-1}+1$$
etc. 
