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

Question

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!

Answer 1

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:

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

Answer 2

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

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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.