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There is a little problem that I worked on, but still wasn't able to prove it completely.
I would be really gratefull if you helped me with it a bit.

Suppose: the sequence k: $\{\frac{a_n}{b_n}\}$ is monotone and $b_n>0$
Prove: that the sequence m: $\{\frac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n}\}$ is monotone as well.

I've been able to proof that:
-if m is increasing $\rightarrow$ $\frac{a_1+a_2+...+a_{n-1}}{b_1+b_2+...+b_{n-1}}\lt\frac{a_n}{b_n}$


For $n=1$ $\land$ k $\nearrow$ you get an easy proof:
$$\frac{a_n}{b_n}\lt\frac{a_{n+1}}{b_{n+1}} $$ $$\frac{a_1}{b_1}\lt\frac{a_{2}}{b_{2}} $$$${a_1}{b_2}\lt{a_{2}}{b_{1}} $$

If m is $\nearrow$ then:$$\frac{a_1}{b_1}\lt\frac{a_1+a_{2}}{b_1+b_{2}}$$$${a_1}(b_1+b_2)\lt({a_1+a_{2})}{b_{1}} $$$${a_1}b_1+a_1b_2\lt{a_1}b_1+a_1b_{2} $$$$a_1b_2\lt{a_2}{b_1} $$$$\downarrow$$ $$\text{Which is the thing that we assumed} $$

So now if you prove it for $n\text{ } \land{n+1}$ you get a full proof. (At least for an increasing sequence, but I assume the proof for a descending one will be quite simillar)

Also do you need to prove that if k is $\nearrow$ than m is $\nearrow$ as well?
If yes, how would you do it?

$$\text{Looking for your suggestions and thank you in advance!}$$

George
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2 Answers2

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One thing you are missing is the "mediant" inequality. The version I remember is: with all letters positive, if $ \frac{p}{q} < \frac{r}{s} $ then $$ \frac{p}{q}< \frac{p+r}{q+s} < \frac{r}{s} $$

As they may be allowing negative numerators, care must be used in that case.

Will Jagy
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Without loss of generality we can assume that $(a_n/b_n)$ is increasing (otherwise replace $a_n$ by $-a_n$). We want to show that $$ \frac{a_1+a_2+\cdots+a_n}{b_1+b_2+\cdots+b_n} \le \frac{a_1+a_2+\cdots+a_n+a_{n+1}}{b_1+b_2+\cdots+b_n+b_{n+1}} $$ which is equivalent to $$ b_{n+1}(a_1+a_2+\cdots+a_n) \le a_{n+1}(b_1+b_2+\cdots+b_n) $$ or $$ \frac{a_1+a_2+\cdots+a_n}{b_1+b_2+\cdots+b_n} \le \frac{a_{n+1}}{b_{n+1}} \, . $$

And that follows from the generalized mediant inequality, since $$ \frac{a_1+a_2+\cdots+a_n}{b_1+b_2+\cdots+b_n} \le \max_i \frac{a_i}{b_i} = \frac{a_n}{b_n} \le \frac{a_{n+1}}{b_{n+1}} $$

Martin R
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    Alternatively, one can avoid any explicit references to mediants by noting $$i \leq n \implies \frac{a_i}{b_i} \leq \frac{a_{n+1}}{b_{n+1}} \implies b_{n+1}a_i \leq a_{n+1}b_i.$$ (aside from that, my solution was going to be identical, so I hope you don't mind it as a comment) – Brian Moehring Jun 05 '22 at 20:01
  • @BrianMoehring: Yes, that is even simpler, you can still post it as an answer! – Martin R Jun 05 '22 at 20:04
  • Thanks for the answer, now I get it! – George Jun 05 '22 at 20:12