Calculus of a Single Variable, Chapter 9, 9.1, Section 9.1, Problem 54

Let write the first few terms of the sequence:
a_1=3/3=1
a_2=6/4=3/2
a_3=9/5
a_4=12/6=2
We can see that the first four terms are increasing so it is possible that the whole sequence is monotonically increasing. To prove that, we need to check that
a_n (3n)/(n+2)<(3(n+1))/(n+1+2)
(3n)/(n+2)<(3n+3)/(n+3)
Now we multiply the whole inequality by (n+2)(n+3). We can do that because n>0 and thus (n+2)(n+3)>0.
3n^2+9n<3n^2+6n+3n+6
3n^2-3n^2+9n-9n<6
0<6
Since zero is indeed less than six we can conclude that a_nBecause the sequence is monotonically increasing it is bounded from below i.e. 1<=a_n, forall n in NN.
Let us now prove that the sequence is bounded from above by 3 i.e.
a_n<3
(3n)/(n+2)<3
Now we multiply by n+2. We can do that because n>0 and thus n+2>0.
3n<3n+6
0<6
Since zero is indeed less than six we can conclude that a_n<3, forall n in NN. Therefore, the sequence is bounded from both below and above i.e.
1leq a_n<3
The image below shows first 50 terms of the sequence. Both monotonicity and boundedness can clearly be seen on the image.