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.

• Looks good to me. A slight shortcut at the last step: $$\frac{n+2}{3n+1}=\frac{n+\cfrac{1}{3}+\cfrac{5}{3}}{3n+1}=\frac{1}{3}+\frac{5}{3(3n+1)}$$ – dxiv Jan 6 '17 at 18:56
• I think a better decomposition of the fraction for this purpose is $\frac13\left(1+\frac4{3n(3n+1)}+\frac5{3n}\right)$ – robjohn Jan 6 '17 at 21:25

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!

Let $$a_n=\frac{n^2+2n+1}{3n^2+n}$$ Then $$a_{n+1}-a_n=\frac{-5n^2-11n-4}{(3n^2+7n+4)(3n^2+n)}<0$$

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 $\frac{1}{a_n}$?

• It's decreasing. Nice. – lmc Jan 6 '17 at 19:11

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.

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.