Find the floor of the product of a series

anemone

MHB POTW Director
Staff member
Evaluate $\left|\dfrac{2017}{2013}\times \dfrac{2009}{2005}\times \cdots \times \dfrac{49}{45}\times \dfrac{41}{37}\times\dfrac{33}{29} \right|$.

MarkFL

Staff member
Re: Find the absolute value

Using W|A, I find the product is:

83836150565816858292640911206036394563843681031859
52356168446790030435481142305738272797214023758253
70545227854661976992847614540194807590851534324428
04393499734248743073514379606568657879433625129913
6361634674742694324903259277229258753

divided by

97080688590435020901312257449744872243043749596819
59388449562938368617693451031135430220537455405367
99055475980516651749335764543619797829673412762497
25762328646784869776620872718676646990400310128382
088560961232371053400465434017358045

anemone

MHB POTW Director
Staff member
Re: Find the absolute value

I am sorry, MarkFL!

I meant to say, find the floor of the expression, i.e:

$\left\lfloor\dfrac{2017}{2013}\times \dfrac{2009}{2005}\times \cdots \times \dfrac{49}{45}\times \dfrac{41}{37}\times\dfrac{33}{29} \right\rfloor$.

MarkFL

Staff member
Re: Find the absolute value

I am sorry, MarkFL!

I meant to say, find the floor of the expression, i.e:

$\left\lfloor\dfrac{2017}{2013}\times \dfrac{2009}{2005}\times \cdots \times \dfrac{49}{45}\times \dfrac{41}{37}\times\dfrac{33}{29} \right\rfloor$.
Yeah right...you are just trying to get to the 1,000 post mark more quickly...you can't fool me.

But, it's good because this helps me get to the 5,000 post mark too.

Seriously though, I kind of thought that's what you meant, but I ran with it anyway.

anemone

MHB POTW Director
Staff member
Re: Find the absolute value

LOL! Though I desperately wanted to hit the 1,000 post mark as soon as possible, I would never post something unintelligible for our members. I don't know why I kept making silly mistakes at our site since the last few days. Perhaps I need to drink more coffee...

mente oscura

Well-known member
Evaluate $\left|\dfrac{2017}{2013}\times \dfrac{2009}{2005}\times \cdots \times \dfrac{49}{45}\times \dfrac{41}{37}\times\dfrac{33}{29} \right|$.
Hello.

$$\displaystyle\prod_{n=1}^{249} (1+\dfrac{1}{2n+5+\frac{1}{4}}) \approx{}8.635718574$$

Regards.

Edit:

At this time:

Rate problem=8

Post anemone=992

992 post+8 post=1000 post. Congrats.

Last edited:

Opalg

MHB Oldtimer
Staff member
Evaluate $\left\lfloor\dfrac{2017}{2013}\times \dfrac{2009}{2005}\times \cdots \times \dfrac{49}{45}\times \dfrac{41}{37}\times\dfrac{33}{29} \right\rfloor$.
I used Stirling's formula $n! \approx \sqrt{2\pi n}(n/e)^n$ to find that $$\frac{2 \cdot 4 \cdot 6\ \cdots\ (2n)\quad }{1 \cdot 3 \cdot 5 \cdots (2n-1)} = \frac{2^2 \cdot 4^2 \cdot 6^2 \cdots (2n)^2}{(2n)!} = \frac{2^{2n}(n!)^2}{(2n)!} \approx \frac{2^{2n}\cdot 2\pi n\cdot n^{2n}\cdot e^{2n}}{2\sqrt{\pi n}\cdot(2n)^{2n}\cdot e^{2n}} = \sqrt{\pi n}.$$ The approximation is correct up to a factor of the order of $1/n$. So if $n$ is around $250$ say, then the approximation should be correct up to a couple of decimal places. Then \begin{aligned}\frac{33}{29} \cdot \frac{41}{37} \cdot \frac{49}{45} \cdots \frac{2017}{2013} &< \frac{32}{28} \cdot \frac{40}{36} \cdot \frac{48}{44} \cdots \frac{2016}{2012} \\ &= \frac{8}{7} \cdot \frac{10}{9} \cdot \frac{12}{11} \cdots \frac{504}{503} = \frac{1\cdot 3 \cdot 5}{2\cdot 4\cdot 6} \cdot \frac{2 \cdot 4 \cdot 6\cdots504}{1 \cdot 3 \cdot 5 \cdots 503} \approx \frac{15\sqrt{252\pi}}{48} \approx 8.79.\end{aligned} To get an inequality the other way round, use Stirling's formula again to find that $$\frac{3 \cdot 5 \cdot 7 \cdots (2n+1)}{2 \cdot 4 \cdot 6\ \cdots\ (2n)\quad } = \frac{(2n+1)!}{\bigl((2n)!\bigr)^2} \approx \frac{2n+1}{\sqrt{\pi n}}.$$ Then \begin{aligned}\frac{33}{29} \cdot \frac{41}{37} \cdot \frac{49}{45} \cdots \frac{2017}{2013} &> \frac{36}{32} \cdot \frac{44}{40} \cdot \frac{52}{48} \cdots \frac{2020}{2016} \\ &= \frac{9}{8} \cdot \frac{11}{10} \cdot \frac{13}{12} \cdots \frac{505}{504} = \frac{2\cdot 4\cdot 6}{3\cdot 5 \cdot 7} \cdot \frac{505!}{(504!)^2} \approx \frac{48\cdot 505}{105\sqrt{252\pi}} \approx 8.20.\end{aligned} Thus $$\displaystyle \frac{33}{29} \cdot \frac{41}{37} \cdot \frac{49}{45} \cdots \frac{2017}{2013}$$ lies between $8.2$ and $8.8$ (in agreement with what mente oscura found), and so its integer part is 8.

Last edited:

anemone

MHB POTW Director
Staff member
Hello.

$$\displaystyle\prod_{n=1}^{249} (1+\dfrac{1}{2n+5+\frac{1}{4}}) \approx{}8.635718574$$

Regards.

Edit:

At this time:

Rate problem=8

Post anemone=992

992 post+8 post=1000 post. Congrats.
Thanks, mente oscura for participating...and I will accept that answer (yes, it is correct but you haven't convinced me how you gotten the approximate value for that product and hence determined its floor value), because you are nevertheless a good member at our site.

Based on your other brilliant calculation about how I gotten the 1k posts, I want to thank you too and also the congratulations note as well. Hehehe... By the way, I am very close to making my 1000th post!

I used Stirling's formula $n! \approx \sqrt{2\pi n}(n/e)^n$ to find that $$\frac{2 \cdot 4 \cdot 6\ \cdots\ (2n)\quad }{1 \cdot 3 \cdot 5 \cdots (2n-1)} = \frac{2^2 \cdot 4^2 \cdot 6^2 \cdots (2n)^2}{(2n)!} = \frac{2^{2n}(n!)^2}{(2n)!} \approx \frac{2^{2n}\cdot 2\pi n\cdot n^{2n}\cdot e^{2n}}{2\sqrt{\pi n}\cdot(2n)^{2n}\cdot e^{2n}} = \sqrt{\pi n}.$$ The approximation is correct up to a factor of the order of $1/n$. So if $n$ is around $250$ say, then the approximation should be correct up to a couple of decimal places. Then \begin{aligned}\frac{33}{29} \cdot \frac{41}{37} \cdot \frac{49}{45} \cdots \frac{2017}{2013} &< \frac{32}{28} \cdot \frac{40}{36} \cdot \frac{48}{44} \cdots \frac{2016}{2012} \\ &= \frac{8}{7} \cdot \frac{10}{9} \cdot \frac{12}{11} \cdots \frac{504}{503} = \frac{1\cdot 3 \cdot 5}{2\cdot 4\cdot 6} \cdot \frac{2 \cdot 4 \cdot 6\cdots504}{1 \cdot 3 \cdot 5 \cdots 503} \approx \frac{15\sqrt{252\pi}}{48} \approx 8.79.\end{aligned} To get an inequality the other way round, use Stirling's formula again to find that $$\frac{3 \cdot 5 \cdot 7 \cdots (2n+1)}{2 \cdot 4 \cdot 6\ \cdots\ (2n)\quad } = \frac{(2n+1)!}{\bigl((2n)!\bigr)^2} \approx \frac{2n+1}{\sqrt{\pi n}}.$$ Then \begin{aligned}\frac{33}{29} \cdot \frac{41}{37} \cdot \frac{49}{45} \cdots \frac{2017}{2013} &> \frac{36}{32} \cdot \frac{44}{40} \cdot \frac{52}{48} \cdots \frac{2020}{2016} \\ &= \frac{9}{8} \cdot \frac{11}{10} \cdot \frac{13}{12} \cdots \frac{505}{504} = \frac{2\cdot 4\cdot 6}{3\cdot 5 \cdot 7} \cdot \frac{505!}{(504!)^2} \approx \frac{48\cdot 505}{105\sqrt{252\pi}} \approx 8.20.\end{aligned} Thus $$\displaystyle \frac{33}{29} \cdot \frac{41}{37} \cdot \frac{49}{45} \cdots \frac{2017}{2013}$$ lies between $8.2$ and $8.8$ (in agreement with what mente oscura found), and so its integer part is 8.
Bravo, Opalg! Just so you know, the method that I wanted to share (that is a method provided by other who is apparently good at math) here is quite similar to yours!

Solution suggested by other, which I improved it for a bit:

First we let
$m=\dfrac{2017}{2013}\times \dfrac{2009}{2005}\times \cdots \times \dfrac{49}{45}\times \dfrac{41}{37}\times\dfrac{33}{29}$

Observe that if we apply the AM-GM inequality to the terms $n$ and $n+8$, we have:

$\dfrac{n+(n+8)}{2}\ge \sqrt{n(n+8)}$

This simplifies to $n(n+8) < (n+4)^2$, since $n \ne n+8$.

Rearranging it a bit we have that $\dfrac{n+8}{n+4}<\dfrac{n+4}{n}$ and numerically this tells us

$\dfrac{2017}{2013}<\dfrac{2013}{2009}$

$\dfrac{2009}{2005}<\dfrac{2005}{2001}$

$\;\;\;\;\;\;\vdots$

$\dfrac{41}{37}<\dfrac{37}{33}$

$\dfrac{33}{29}<\dfrac{29}{25}$

Multiplying them out gives

$\dfrac{2017}{2013}\times \dfrac{2009}{2005}\times \cdots \times \dfrac{41}{37}\times\dfrac{33}{29}<\dfrac{2013}{2017}\times \dfrac{2005}{2001}\times \cdots \times \dfrac{45}{41}\times\dfrac{37}{33}$

or $m<\dfrac{2013}{2017}\times \dfrac{2005}{1999}\times \cdots \times \dfrac{53}{49}\times \dfrac{45}{41}\times\dfrac{37}{33}$

Next, if we multiply both sides of the inequality with $m$, that yields

$m^2<\dfrac{2017}{25}$

$m<\sqrt{\dfrac{2017}{25}}<8.9822046$

By the similar token, we also have $\dfrac{n+4}{n} >\dfrac{n+8}{n+4}$,

$\dfrac{2007}{2013}\>\dfrac{2025}{2017}$

$\dfrac{2009}{2005}> \dfrac{2013}{2009}$

$\;\;\;\;\;\;\vdots$

$\dfrac{41}{37}\ge \dfrac{45}{41}$

$\dfrac{33}{29}\ge \dfrac{37}{33}$

Multiplying them out gives

$m>\dfrac{2025}{2017}\times \dfrac{2013}{2009}\times \cdots \times \dfrac{37}{33}\times\dfrac{45}{41}$

Multiply both sides by $m$ gives

$m^2>\dfrac{2025}{29}$

$m>\sqrt{\dfrac{2025}{29}}>8.35629$

By combining the two inequalities of $m$, we have $8.35629<m<8.9822046$, so we have the floor of $m$ as 8.