Answer
$a_{n}=\frac{n+2}{n+3}$
Or:
$a_{n}=\frac{3+(n-1)}{4+(n-1)}$
Work Step by Step
We are given:
$\frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}$ ...
We notice that the numerator and denominator both increase by $1$ between terms, but start with different values. We find the pattern:
$a_1=\frac{1+2}{1+3}$
$a_2=\frac{2+2}{2+3}$
$a_3=\frac{3+2}{3+3}$
$a_4=\frac{4+2}{4+3}$
Therefore:
$a_{n}=\frac{n+2}{n+3}$
Or:
$a_{n}=\frac{3+(n-1)}{4+(n-1)}$
