Let ω(n) be the number of distinct prime factors of the integer n, with us defining ω(1)=0 (see, for example, here). What, then, is the value of the series }}{n^2})
?
How about }}{n^4})
?
Note first that if a and b are mutually prime, then =\omega(a)+\omega(b))
, and thus }=2^{\omega(a)}2^{\omega(b)})
.
Now the fundamental theorem of arithmetic tells us that every integer n>1 has a unique prime factorization 
, where the p1, p2, …, pk are distinct primes, and the m1, m2, …, mk are positive integers (note that k=ω(n)). Then, for n>1, =\omega(p_1^m_1)+\omega(p_2^m_2)+\cdots+\omega(p_k^m_k))
and }=2^{\omega(p_1^m_1)}2^{\omega(p_2^m_2)}\cdots2^{\omega(p_k^m_k)})
.
We see then that the prime product
}}{p^{k\cdot{s}}}\right])
, when expanded into a sum, will contain terms whose numerator is of the form }2^{\omega(p_2^k_2)}\cdots2^{\omega(p_n^k_n)})
, with corresponding denominator ^s)
, with each such combination appearing once. But these are just })
and 
, with each positive integer n appearing once and only once. Thus
}}{n^s}=\prod_{p\;\text{prime}}\left[\sum_{k=0}^{\infty}\frac{2^{\omega(p^k)}}{p^{k\cdot{s}}}\right])
and as }=2^0=1)
, and }=2^1=2)
for k>0, we have
}}{n^s}=\prod_{p\;\text{prime}}\left[1+\sum_{k=1}^{\infty}\frac{2}{p^{ks}}\right])
Now, via the geometric series,
so
and thus
}}{n^s}=\prod_{p\;\text{prime}}\left(\frac{1+p^{-s}}{1-p^{-s}}\right))
.
Now, we showed previously that =\prod_{p\;\text{prime}}\frac{1}{1-p^{-s}})
, and so
}}{n^s}=\zeta(s)\cdot\prod_{p\;\text{prime}}\left(1+p^{-s}\right))
.
To find this last product, note that =\frac{1-x^2}{1-x})
. Thus
=\frac{\prod_{p\;\text{prime}}\left(1-p^{-2s}\right)}{\prod_{p\;\text{prime}}\left(1-p^{-s}\right)})
, and again using , we see that
=\frac{\zeta(s)}{\zeta(2s)})
, and so
}}{n^s}=\frac{\zeta^2(s)}{\zeta(2s)})
.
For s=2, this then answers our original question:
}}{n^2}=\frac{\zeta^2(2)}{\zeta(4)})
, and using =\frac{\pi^2}{6})
and =\frac{\pi^4}{90})
(see here), we find
}}{n^2}=\frac{\left(\frac{\pi^2}{6}\right)^2}{\frac{\pi^4}{90}}=\frac{5}{2})
.
Similarly, =\frac{(-1)^{5}2^{7}\pi^{8}}{(8)!}B_{8}=\frac{\pi^8}{9450})
(see here), and so
}}{n^4}=\frac{\zeta^2(4)}{\zeta(8)}=\frac{7}{6})
.
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Tags: Geometric Series, Math, Monday Math, Prime Factors, Prime Numbers, Riemann Zeta Function
This entry was posted on February 23, 2009 at 12:05 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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