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 <a_(n+1)`

`(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_n<a_(n+1),` `forall n in NN.`Therefore, the **sequence is** **monotonically increasing**.

Because 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. ` `