How to prove if function is strictly monotonic increasing

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Solution 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.

Solution 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.


[ 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.

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Updated on August 16, 2022

Comments

  • Michael
    Michael about 1 year

    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?

    • Samiron
      Samiron about 7 years
      you may try to use the fact that a sequence $a_n$ is increasing iff the sequence $b_n = (5/3)a_n -1 $is increasing.
    • amWhy
      amWhy about 7 years
      Not a good idea to ask for "the best way...", "easiest way..." Those matters are in the eye of the beholder. Besides, if you get someone willing to offer a way, you should be grateful to know one way, since you clearly haven't shown any attempt at any proof whatsoever.
    • amWhy
      amWhy about 7 years
      What is the function? I'm not going to chase links to find it. It's easy enough for you to type it into your question, if the question matters to you.
  • Michael
    Michael about 7 years
    Could you please add more details.
  • amWhy
    amWhy about 7 years
    @Michael Beggars can't be choosers. Please ask specific questions. Have you tried to prove by induction?
  • amWhy
    amWhy about 7 years
    @Michael: the question of the hour is "Could you provide more details, in your post? If you ask a question with no details, no effort, no explanation, you have no right to insist that others give you any more than they graciously gave you, given the little you provide.
  • wythagoras
    wythagoras about 7 years
    Did you mean $a_1>a_0$?
  • Bargabbiati
    Bargabbiati about 7 years
    You're right, I was too vague.