How to prove if function is strictly monotonic increasing

Question

I have this function and I have to prove that this function is strictly monotonic increasing

This is the equation: $$a_n = \frac{3n-7}{8+5n}$$

What would be the best way to do this?

Answer 1

Hint: induction. $a_1>a_0$, in fact $-\frac 7 8<-\frac 4 {13}$. Than given that $a_n>a_{n-1}$ let's show that $a_{n+1}>a_n$. You have $a_{n+1}=\frac {3(n+1)-7} {8+5(n+1)}=\frac {3n-4} {13+5n}, a_n={\frac {3n-7} {8+5n}}$, so you have to show that $\frac {3n-4} {13+5n}>\frac {3n-7} {8+5n} \iff(3n-4)(5n+8)>(3n-7)(13+5n)\iff 15n^2+4n-32>15n^2+4n-91$

and it's done.

Answer 2

Write $a_n$ as: $$a_n=\frac{15 n - 35}{15 n + 24} = 1 - \frac{59}{15n + 24}$$

then notice that:

• $15n+24\;$ is strictly positive and increasing, thus

• $\frac{59}{15n + 24}\;$ is decreasing, thus

• $-\frac{59}{15n + 24}\;$ is increasing, thus

• $a_n=-\frac{59}{15n + 24} + 1\;$ is increasing.

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[ EDIT ]    The above makes use of the known (and otherwise easy to prove) properties:

• if $f(x) \gt 0$ for all $x$, then $f(x)$ is [strictly] increasing $\iff$ $\frac{1}{f(x)}$ is [strictly] decreasing;

• $f(x)$ is [strictly] increasing $\iff$ $-f(x)$ is [strictly] decreasing.