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I have two progressions: $$\begin{align}P_1&: xy, xy + (x+2)(y+2), xy + (x+2)(y+2) + (x+4)(y+4),\dots, \sum_1^i (x+2(j-1))(y+2(j-1))\end{align}$$ $$\begin{align}P_2&: (x+2)(y+2), (x+4)(y+4),\dots, (x + 2i)(y+2i)\end{align}$$

$x, y$ are natural numbers.

I just want to know if the ratio $P_1(i):P_2(i)$ is an increasing function. It appears to be. But how to prove it?

I would like to add that $x > 3$, $y > 3$ are odd numbers.

KingLogic
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Marina
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  • It depends on the choice of $x$ and $y$ – Raffaele Sep 20 '20 at 20:22
  • Care to elaborate? I do not see, how. It appears, it is always increasing in my experiments.Can you give an example, when it is not increasing? – Marina Sep 20 '20 at 20:23
  • For $x=2,y=3$ I got $0.0733333, 0.0553288, 0.0446208, 0.0374708, 0.0323389, 0.0284676, 0.0254389, 0.0230022, 0.0209979, 0.0193195$ – Raffaele Sep 20 '20 at 20:35
  • Thank you. What do you think, is it possible to find x, y where the ratio not increases if x, y are odd numbers x >3 y > 3? It would be great to know. – Marina Sep 20 '20 at 20:38
  • I calculated for $x=2, y=3$, and I have increasing sequence:0.3000000 0.6190476 0.9444444 1.2727273 1.6025641 1.9333333 2.2647059 2.5964912 2.9285714 3.2608696 3.5933333 3.9259259 4.2586207 4.5913978 4.9242424 5.2571429 5.5900901 – Marina Sep 20 '20 at 21:07

1 Answers1

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Term $k$ of $P_1$ is $$[x+2(k-1)][y+2(k-1)]=xy+2(k-1)(x+y)+4(k-1)^2$$ Summing by terms from $k=1..n$ we get $$P_1(n)=nxy+n(n-1)(x+y)+\frac23(2n-1)n(n-1)$$ Meanwhile $$P_2(n)=(x+2n)(y+2n)=xy+2n(x+y)+4n^2$$

Therefore as $n \to \infty$ then $$\frac{P_1(n)}{P_2(n)} \to \frac14(x+y)+\frac13 n$$ which continues increasing indefinitely for all values of $x, y$.

sammy gerbil
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