I came across an exercise where one is supposed to find the bounds of the following sequence, given by $$a_n=\frac{n+2}{n-3}$$ where n is a Natural number, which starts at n=1 and goes to infinity.
We can check that $a_{n+1}-a_n<0$ and $a_1=-1,5$. Therefore, can we say that the sequence is decreasing and that is bounded from above? And, however, how can we know if it is also bounded from below? And how do we deal with $a_3=\frac{5}{3-3}$.
The sequence is defined in $\Bbb{N}$ \ {3}.
Let's study the sequence before and after the "breaking point" n=3 . For $n \lt 3$ we have just to check a finite number of $a_n$, to be precise we have $a_1=-3/2$ and $a_2=-4$.
Then we have to check the sequence for $n\gt 3$.
For $n\gt3$ as you said it's true that $a_{n+1}\lt a_n$ and is also true that $1\lt\frac{n+2}{n-3}=a_n$ .Then for $n\gt3$ we can affirm that $a_4=7/2$ is the above bound, and 1 is the lower bound.
$EDIT$ to affirm that 1 is the lower bound you also need to show that $\lim_{n \to \infty}\frac{n+2}{n-3}=1_+$, in other words that limit go to 1 from above.
Now we have to find the bound of all the sequence, so we check also for $a_1$ and $a_2$, and we conclude that the lower bound of the sequence is $a_2=-4$.
The not trivial point of how we got the inequality $1\lt\frac{n+2}{n-3}=a_n$ is that we just used some function study, of the function $f(x)=\frac{x+2}{x-3}$, and compute $\lim_{x\to \infty}{\frac{x+2}{x-3}}$ .
Hope this helps,
best regards,
Andrea
P.S. The study of the function $f(x)=\frac{x+2}{x-3}$ also tell you why $n=3$ is a "breaking point".