(n^3+3n)^2/(n^6-64)

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Source: IMC 2019 Day 1 P1

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#1 h Jul 31, 2019, 7:41 AM • 5 Y

Y by Mathuzb, Davrbek, Adventure10, Mango247, mathematicsy

Evaluate the product
$\prod_{n=3}^{\infty} \frac{(n^3+3n)^2}{n^6-64}.$
Proposed by Orif Ibrogimov, ETH Zurich and National University of Uzbekistan and Karen Keryan, Yerevan State University and American University of Armenia, Yerevan

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TuZo

#2 h Jul 31, 2019, 8:27 AM • 2 Y

Y by Adventure10, Mango247

Solution:
We try to get telescoping product: $P=\prod\limits_{n=1}^{m}{\frac{{{({{n}^{3}}+3n)}^{2}}}{{{n}^{6}}-64}}=\prod\limits_{n=1}^{m}{\frac{{{n}^{2}}{{({{n}^{2}}+3)}^{2}}}{(n-2)(n+2)({{(n+1)}^{2}}+3)({{(n-1)}^{2}}+3)}}=\prod\limits_{n=1}^{m}{\frac{n}{n-2}}\prod\limits_{n=1}^{m}{\frac{n}{n+2}\prod\limits_{n=1}^{m}{\frac{{{n}^{2}}+3}{{{(n+1)}^{2}}+3}}}\prod\limits_{n=1}^{m}{\frac{{{n}^{2}}+3}{{{(n-1)}^{2}}+3}}$
We get: $P=\left( \prod\limits_{n=1}^{m}{\frac{n}{n-2}} \right)\frac{2}{(m+1)(m+2)}\frac{28}{{{(m+1)}^{2}}+3}\frac{{{m}^{2}}+3}{12}$ , but the first product do not exist for the $n=2$ , we must change something!
The correct problem maybe is this: $\prod_{n=3}^{\infty} \frac{(n^3+3n)^2}{n^6-64}.$

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#3 h Jul 31, 2019, 10:03 AM • 2 Y

Y by Adventure10, Mango247

@above: you're right, the index $n$ in the original problem runs from $3$ to $\infty$ . See https://www.imc-math.org.uk/?year=2019&section=problems&item=prob1q

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#4 h Jul 31, 2019, 10:08 AM • 2 Y

Y by Adventure10, Mango247

In this case $P=\frac{m(m-1)}{2}\frac{2}{(m+1)(m+2)}\frac{12}{{{(m+1)}^{2}}+3}\frac{{{m}^{2}}+3}{7}\to \frac{12}{7}$

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#5 h Sep 19, 2020, 4:21 AM • 2 Y

Y by Friend-from-Taiwan, animath_314159

No. Answer is $72/7$

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#6 h Sep 19, 2020, 4:27 AM • 1 Y

Y by Mango247

https://math.stackexchange.com/questions/3830222/product-of-fracr3-3r2r6-64-to-infinity?noredirect=1#comment7899433_3830222. It telescopes pretty nicely

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denery

#7 h Dec 11, 2020, 10:55 AM • 3 Y

Y by Mango247, Mango247, Mango247

looks pretty easy but is exceptionally hard

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#8 h Jan 30, 2021, 5:56 PM • 1 Y

Y by jolynefag

wait , are you guys serious ?
this is easy problem. i got 72/7
there are just cheap tricks to be done which i found within very less time maybe 5-10 minutes.
not here to depress you guys.
but, i don't know why i as a normal math student solved it which others are finding exceptionally hard.

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#9 h Jan 30, 2021, 7:01 PM

Y by

alexheinis wrote:

@above: you're right, the index $n$ in the original problem runs from $3$ to $\infty$ . See https://www.imc-math.org.uk/?year=2019&section=problems&item=prob1q

Why you must comment, when there are no any mistake?! :mad:

:mad:

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#10 h Jul 25, 2024, 10:37 AM

Y by

denery wrote:

looks pretty easy but is exceptionally hard

It's not as hard as it seems, but it's quite tricky if you didn't follow its pattern correctly/you are careless

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#11 h Dec 2, 2024, 10:51 PM

Y by

$\begin{align*} \prod_{n=3}^\infty \frac{(n^3+3n)^2}{n^6-64} & = \lim_{N\to\infty}\left( \prod_{n=3}^N\frac{(n^3+3n)^2}{n^6-64}\right)\\ & = \lim_{N\to\infty}\left( \prod_{n=3}^N\frac{n^2(n^2+3)^2}{(n^3-8)(n^3+8)}\right)\\ & = \lim_{N\to\infty}\left( \prod_{n=3}^N\frac{n^2(n^2+3)^2}{(n-2)(n^2+2n+4)(n+2)(n^2-2n+4)}\right)\\ & = \lim_{N\to\infty}\left( \prod_{n=3}^N\frac{n}{n-2}\cdot\frac{n^2+3}{(n+1)^2+3}\cdot\frac{n}{n+2}\cdot\frac{n^2+3}{(n-1)^2+3}\cdot\right)\\ & = \lim_{N\to\infty}\left( \frac{N(N-1)}{2}\cdot\frac{12}{(N+1)^2+3}\cdot\frac{12}{(N+1)(N+2)}\cdot\frac{N^2+3}{7}\right)\\ & =\frac{72}{7} \lim_{N\to\infty}\left( \frac{N(N-1)}{1}\cdot\frac{1}{(N+1)^2+3}\cdot\frac{1}{(N+1)(N+2)}\cdot\frac{N^2+3}{1}\right)\\ & =\frac{72}{7} \\ \end{align*}$

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#12 h Apr 4, 2025, 12:16 AM

Y by

denery wrote:

looks pretty easy but is exceptionally hard

wym exceptionally hard? ahahahha

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