<s>[INST] 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.
 [/INST] 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.
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