When trying to spot a pattern in a problem like this, it’s often helpful not to reduce the fractions. In this case every numerator is initially $3$, and the fractions are:
Since all of the numerators are $3$, we can concentrate on the denominators. These are:
You might notice that each is $1$ less than perfect square. If you do, you’ll quickly conjecture that
Alternatively, you might try factoring them:
This would suggest that
In either case you can then prove the result by induction on $n$.
Of course if you happen to see JasonM’s approach, you bypass the pattern-recognition stage and need no induction argument.