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...") |
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+ | ==Problem== | ||
+ | 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 | ||
+ | |||
+ | <cmath> \prod_{i=1}^{\infty}\frac{c_i^2}{c_i^2-1} </cmath> | ||
+ | |||
+ | (Hint: <math> \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6} </math>) | ||
+ | |||
==Solution== | ==Solution== | ||
Latest revision as of 18:37, 14 January 2020
Problem
Let denote the
composite integer so that
. Compute
(Hint: )
Solution
Denote as
. To begin, let
=
.
Let's notice that:
And that:
Notice that all the fractions, except and
will cancel out with their reciprocals from the next term(i.e
cancels with
,
with
).
Therfore:
So we have now proven that . We will save this for later. For the second part, we will use the famous identity (discovered by Euler) that:
.
Plugging in , we see that:
.
Therefore, we can split into two infinite products, one of prime numbers, and one of composite numbers
Noticing that:
We can conclude that:
Therefore, the answer is ,