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Difference between revisions of "2006 SMT/Team Problems/Problem 15"

(Created page with "==Solution== Denote <math>\prod_{i=0}^{\infty} \frac{c_i^{2}}{c_i^{2}-1}</math> as <math>\prod_{composite} \frac{c^2}{c^2-1}</math>. To begin, let <math>S</math> = <math>\pro...")
 
m (→‎Solution)
 
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==Problem==
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Let <math> c_i </math> denote the <math> i\text{th} </math> composite integer so that <math> \{c_i\}=4, 6, 8, 9\cdots </math>. Compute
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<cmath> \prod_{i=1}^{\infty}\frac{c_i^2}{c_i^2-1} </cmath>
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(Hint: <math> \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6} </math>)
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==Solution==
 
==Solution==
  

Latest revision as of 18:37, 14 January 2020

Problem

Let $c_i$ denote the $i\text{th}$ composite integer so that $\{c_i\}=4, 6, 8, 9\cdots$. Compute

\[\prod_{i=1}^{\infty}\frac{c_i^2}{c_i^2-1}\]

(Hint: $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$)

Solution

Denote $\prod_{i=0}^{\infty} \frac{c_i^{2}}{c_i^{2}-1}$ as $\prod_{composite} \frac{c^2}{c^2-1}$. To begin, let $S$ = $\prod_{n=2}^{\infty} \frac{n^2}{n^2-1}$.

Let's notice that:

\begin{align*} \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} \frac{n^2-1}{n^2}\\ \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} (1-\frac{1}{n^2})\\ \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} (1-n^{-2})\\ \end{align*}

And that:

\begin{align*} \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} \frac{n^2-1}{n^2}\\ \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^2}\\ \frac{1}{S} &= \frac{1\cdot3}{2\cdot2} \cdot \frac{2\cdot4}{3\cdot3}\cdots \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot\frac{(k-1)\cdot(k+1)}{k\cdot k} \end{align*}

Notice that all the fractions, except $\frac{1}{2}$ and $\frac{k+1}{k}$ will cancel out with their reciprocals from the next term(i.e $\frac{3}{2}$ cancels with $\frac{2}{3}$, $\frac{k}{k-1}$ with $\frac{k-1}{k}$).

Therfore:

\begin{align*} \frac{1}{S} &= \lim_{k\to\infty} \frac{1}{2} \cdot \frac{k+1}{k}\\ \frac{1}{S} &= \lim_{k\to\infty} \frac{1}{2} \left(1+\frac{1}{k}\right)\\ \frac{1}{S} &=\frac{1}{2} \end{align*}

So we have now proven that $\frac{1}{S} = \frac{1}{2}$. We will save this for later. For the second part, we will use the famous identity (discovered by Euler) that: \[\prod_{primes} (1-p^{-s}) = \frac{1}{\xi(s)}\].

Plugging in $s=2$, we see that: \[\prod_{primes} (1-p^{-2}) = \frac{1}{\xi(2)} = \frac{6}{\pi^2}\].

Therefore, we can split $\frac{1}{S}=\frac{1}{2}$ into two infinite products, one of prime numbers, and one of composite numbers \begin{align*} \frac{1}{S} &= \prod_{n=2}^{\infty} (1-n^{-2})\\ \frac{1}{2} &= \prod_{primes} (1-p^{-2}) \cdot \prod_{composite} (1-c^{-2})\\ \frac{1}{2} &= \frac{1}{\xi(2)}\cdot \prod_{composite} (1-c^{-2})\\ \frac{1}{2} &= \frac{6}{\pi^2}\prod_{composite} (1-c^{-2})\\ \frac{\pi^2}{12} &= \prod_{composite} (1-c^{-2}) \end{align*}

Noticing that:

\[\prod_{composite} (1-c^{-2}) = \prod_{composite} (1-\frac{1}{c^2}) = \prod_{composite} \frac{c^2-1}{c^2}\]

We can conclude that: \begin{align*} \frac{\pi^2}{12} &= \prod_{composite} \frac{c^2-1}{c^2}\\ \frac{12}{\pi^2} &= \prod_{composite} \frac{c^2}{c^2-1} \end{align*}

Therefore, the answer is $\boxed{\frac{12}{\pi^2}}$, $\emph{Dividend}$

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