Can $a(n) = \frac{n}{n+1}$ be written recursively?

Question

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

Answer 1

\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*}

Answer 2

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 ($n+1$).

Answer 3

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 $$A_1 = \frac{1}{2}$$

Answer 4

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.

Answer 5

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 $$ \frac1{2-a_n}=a_{n+1}\\ $$