Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing: 
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)}.$$
Let's prove $\frac{n+2}{3n+1}$ is a decreasing sequence.
$$a_n>a_{n+1} \Leftrightarrow \frac{n+2}{3n+1}>\frac{n+3}{3n+4}\Leftrightarrow(n+2)(3n+4)>(n+3)(3n+1)\Leftrightarrow3n^2+10n+8>3n^2+10n+3\Leftrightarrow 8>3$$
So $\frac{n+2}{3n+1}$ is a decreasing sequence and we know that $\frac{1}{n(3n+1)}$ is also decreasing so our given sequence is a decreasing sequence as a sum of $2$ decreasing sequences.

You're solution is fine.  Here is another, perhaps more efficient way forward.
We start with the decomposition in the OP as expressed by
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)} \tag 1$$
Then, we simply note that the first term on the right-hand side of $(1)$ can be written as
$$\frac{n+2}{3n+1}=\frac13 \frac{3n+6}{3n+1}=\frac13 \left(1+ \frac{5}{3n+1}\right) \tag 2$$
from which we see by inspection that $\frac{n+2}{3n+1}$ is decreasing.  And we are done!
