Need help with determining if sequences $a_{n}=\frac{n+3}{n+2}$ and $a_{n}=\frac{n-1}{n+1}$ are increasing or decreasing I'm struggling with a problem that asks me to determine whether the following sequences are increasing or decreasing:
a) $a_{n}=\frac{n+3}{n+2}$
b) $a_{n}=\frac{n-1}{n+1}$
For part (a), I tried to find $a_{n+1}-a_{n}$ and simplify it to see if it was positive or negative. After some algebra, I got:
$a_{n+1}-a_{n}=\frac{1}{n+2}$
Since $n+2$ is always positive, $a_{n+1}-a_{n}$ is always positive. Therefore, the sequence is increasing.
For part (b), I followed the same process and got:
$a_{n+1}-a_{n}=-\frac{2}{(n+1)(n+3)}$
Since $(n+1)(n+3)$ is always positive, $a_{n+1}-a_{n}$ is always negative. Therefore, the sequence is decreasing.
Could someone please verify if my answers are correct or not? If I made a mistake, I would appreciate any guidance on how to approach the problem correctly.
Thank you in advance for your help!
 A: Your computations are incorrect. In the first case, $a_{n+1}-a_n=-\frac{1}{(n+1)(n+2)}<0$ so $a_n$ is decreasing. In the second case, $a_{n+1}-a_n=\frac{2}{(n+2)(n+1)}>0$ so the sequence is increasing. Your general reasoning is good though.
Alternatively, since $a_n>0$ for every $n$, you can also compute $\frac{a_{n+1}}{a_n}$. Then:

*

*$a_n$ is increasing if and only if $\frac{a_{n+1}}{a_n}>1$ for every $n$

*$a_n$ is decreasing if and only if $\frac{a_{n+1}}{a_n}<1$ for every $n$
For some sequences, this would lead to easier computations.
Finally, in a), you can also write $a_n=1+\frac{1}{n+2}$ and since $\frac{1}{n+2}$ is decreasing, $a_n$ is decreasing.
A: For first part $$a_n=\frac{n+3}{n+2}=1+\frac{1}{n+2}$$
$n$ increases $\implies$ $a_n$ decreases
For second part $$a_n=\frac{n-1}{n+1}=1-\frac{2}{n+1}$$
$n$ increases $\implies$ $a_n$ increases

This is because in first case when $n$ increases the fraction $\frac{1}{n+2}$ decreases. This means everytime we increase $n$ we add something smaller than the previous value as $\frac{1}{n+2}>\frac{1}{(n+1)+2}$
In second case we're subtracting so the opposite of first case will happen.
