Mrthdrb

Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$

Algebraically it can be written as $$a(n) = fracnn + 1$$

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

beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*

After some further solving, I was able to come up with an answer

It can be written $$A_n + 1 = frac12 - A_n$$ where $$A_1 = frac12$$

Just by playing around with some numbers, I determined a recursive relation to be

$$a_n = fracna_n-1 + 1n+1$$

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 ($n+1$).

Perhaps a bit simpler is to note that
$$
overbrace frac11-a_n ^n+1+1=overbracefrac11-a_n+1^n+2\
$$
solving for $a_n+1$ yields
$$
frac12-a_n=a_n+1\
$$

You can, and in multiple ways. Such as:$$a_n=frac1a_n-1+frac2ntag1$$

or

$$a_n=fracna_n-1+1n+1tag2$$

etc.

equation 1, is simply noting: $$fracn-1n+frac2n=fracn+1n=frac1a_n$$ Where the first fraction in the sum is $a_n-1$

and equation 2, simply notes:

$$n=na_n-1+1$$

etc.