Who's gonna be the topper from our community! RA vs SA

Calculus Level 3

Two infinite products A A and B B are defined as follows:

A = ∏ n = 2 ∞ ( 1 − 1 n 3 ) , B = ∏ n = 1 ∞ ( 1 + 1 n ( n + 1 ) ) . A = \displaystyle\prod\limits_{n=2}^{\infty} \left(1-\frac{1}{n^3}\right), \quad B =\displaystyle\prod\limits_{n=1}^{\infty}\left(1+\frac{1}{n(n+1)}\right).

If A B = m k , \frac{A}{B} = \frac{m}{k}, where m m and k k are relatively prime positive integers, determine 100 m + k 100m+k .


The answer is 103.

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Parth Lohomi by Parth Lohomi , 20, India

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2 solutions

Parth Lohomi
May 24, 2015

We get A B = ∏ i = 2 ∞ ( n − 1 ) ( n 2 + n + 1 ) n 3 3 2 ⋅ ∏ i = 2 ∞ n 2 + n + 1 n ( n + 1 ) = 2 3 ∏ i = 2 ∞ ( n − 1 ) ( n + 1 ) n 2 \dfrac{A}{B} = \dfrac{\displaystyle\prod_{i=2}^{\infty} \dfrac{(n-1)(n^2 + n + 1)}{n^3}}{\dfrac32 \cdot \displaystyle\prod_{i=2}^{\infty} \dfrac{n^2 + n + 1}{n(n+1)}} = \dfrac23 \prod_{i=2}^{\infty} \dfrac{(n-1)(n+1)}{n^2} which telescopes to

2 3 ⋅ 1 2 = 1 3 ⟹ m = 1 , k = 3 ⟹ 100 m + n = 103 \dfrac23 \cdot \dfrac12 = \dfrac13 \implies m=1,k=3\implies 100m+n=\boxed{103} .

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Moderator note:

Almost perfect. Be careful with your notations, because it shows whether you actually pair up the terms or not. We are dealing with infinities. When taking their ratios, you should use lim ⁡ n → ∞ ∏ i = 2 n ( … ) \displaystyle \lim_{n\to \infty} \prod_{i=2}^n (\ldots ) instead of ∏ i = 2 ∞ ( … ) \displaystyle \prod_{i=2}^\infty (\ldots ) . Do you see why?

Bonus question: Prove that A = 1 3 π cosh ⁡ ( 3 2 π ) A = \frac1{3\pi} \cosh\left(\frac {\sqrt3}2 \pi\right) .

Nice question and solution, Parth. In the expressions for A A and B B you may want to change the n n 's to i i 's to be consistent with the product index.

Brian Charlesworth - 6 years ago

Looks like Ramanujan has solved these some 100 years ago: Related link: http://www.imsc.res.in/~rao/ramanujan/collectedpapers/question/q261.htm

chandrasekhar S - 6 years ago

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Yes he is god of series and patterns according to me and such series can be solved by just his fingures.

D K - 2 years, 10 months ago

i got the last expression too bu i don't think 1 − 1 n 2 1-\frac { 1 }{ { n }^{ 2 } } telescopes , can you plz explain it !

hiroto kun - 4 years, 5 months ago

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It's a tough call since he is getting nearly the same marks as I am getting, you see we both are in the same camp , sharing the same room.

But still I think he will get more marks than me.

Ronak Agarwal - 6 years ago

it looks like SA( @Sheshansh Agrawal ) won ! sheshansh-58 ronak-77 !

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