5
$\begingroup$

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.

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

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!

$\endgroup$
4
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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 $\frac{1}{a_n}$?

$\endgroup$
  • $\begingroup$ It's decreasing. Nice. $\endgroup$ – lmc Jan 6 '17 at 19:11
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.