I'm trying to solve the following:
Let $n\in \mathbb N$ and $\{x_n\}$ be a sequence defined by: $$ \{x_n\} = \frac{an + b}{cn+d} $$ Find the relation between $a,b,c,d$ such that $\{x_n\}$ is
- increasing
- decreasing
starting from some index $n_0$.
Since this question is for precalculus level i assumed that $a,b,c,d \in \mathbb R$. Consider the difference between terms:
$$ \require{cancel} x_n - x_{n+1} = \frac{an + b}{cn+d} - \frac{a(n+1) + b}{c(n+1)+d} = \\ = \frac{(an + b)(c(n+1)+d) - (cn+d)(a(n+1) + b)}{(cn+d)(c(n+1)+d)} =\\ = \frac{\cancel{acn^2} + \cancel{acn}+\cancel{adn}+\cancel{bcn}+bc+\cancel{bd}-\cancel{acn^2}-\cancel{adn}-\cancel{acn}-ad-\cancel{bcn}-\cancel{bd}}{(cn+d)(c(n+1)+d)} $$
Nominator reduces to $bc - ad$, so:
$$ x_n - x_{n+1} = \frac{bc-ad}{(cn+d)(c(n+1)+d)} $$
The answer suggests that the sequence increases if $ad > bc$ and decreases if $ad < bc$, which follows directly from the equation above but only with the assumption that denominator is positive.
My question is why shouldn't we bother abound the sign of the denominator in this case?
I don't see how it follows from the denominator:
$$ (cn+d)(cn+c+d) = c^2n^2+c^2n+cdn+ cdn+cd+d^2 = c^2n(n+1) + cd(2n+1)+d^2 $$
From here: $$ \begin{cases} c^2n(n+1) > 0 \\ d^2 > 0 \end{cases} $$
But $cd(2n+1)$ may be less than $0$ which depends on the sign of $c$ and $d$ and therefore denominator be less than $0$. Seems like i'm missing something super obvious.
By derivation we obtain the answer simpler than this method, but in your case we have
If $c>0$ $$x_n - x_{n+1} = \frac{bc-ad}{(cn+d)(c(n+1)+d)}<\frac{bc-ad}{(cn+d)^2}<0$$ when $ad-bc>0$, and sequence in increasing.
If $c>0$ $$x_n - x_{n+1} = \frac{bc-ad}{(cn+d)(c(n+1)+d)}>\frac{bc-ad}{(cn+c+d)^2}>0$$ when $ad-bc<0$, and sequence in decreasing.
If $c<0$ $$x_n - x_{n+1} = \frac{bc-ad}{(cn+d)(c(n+1)+d)}>\frac{bc-ad}{(cn+d)^2}>0$$ when $ad-bc<0$, and sequence in decreasing.
If $c<0$ $$x_n - x_{n+1} = \frac{bc-ad}{(cn+d)(c(n+1)+d)}<\frac{bc-ad}{(cn+c+d)^2}<0$$ when $ad-bc>0$, and sequence in increasing.