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.
 A: Let $$a_n=\frac{n^2+2n+1}{3n^2+n}$$ Then
$$a_{n+1}-a_n=\frac{-5n^2-11n-4}{(3n^2+7n+4)(3n^2+n)}<0$$
A: Your solution looks good.
Another approach could be:
$$a_n=\frac{3n^2+n}{(n+1)^2}=\frac{3(n+1)^2-5n-3}{(n+1)^2}=3-\left[\frac{5n}{(n+1)^2}+\frac{3}{(n+1)^2}\right]=3-\left[\frac{5}{n+2+\frac{1}{n}}+\frac{3}{(n+1)^2}\right]$$
$a_n$ is increasing so what can we conclude about $\frac{1}{a_n}$?
A: HINT  Find the difference (an+1 - an) and study the sign of this difference. If it is positive, the sequence is increasing, otherwise it is decreasing.
A: 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!
A: After breaking up the fraction using Partial Fractions, we see that $\frac3n$ is bigger than $\frac4{3n+1}$, so we give $\frac4{3n}$ of $\frac3n$ to $-\frac4{3n+1}$ to make it positive, but decreasing.
$$
\begin{align}
\frac{n^2+2n+1}{3n^2+n}
&=\frac13\left(1+\frac{5n+3}{3n^2+n}\right)\\
&=\frac13\left(1-\frac4{3n+1}+\frac3n\right)\\
&=\frac13\left(1-\frac4{3n+1}+\frac4{3n}+\frac5{3n}\right)\\
&=\frac13\left(1+\frac4{3n(3n+1)}+\frac5{3n}\right)\tag{1}
\end{align}
$$
For $n\gt0$, each non-constant term in $(1)$ is decreasing.
