- The following links the multiplication of Dirichlet series with the divisibility properties of their coeficients.
- Show that if \(\{ a _ { m } \}\) and \(\{ b _ { k } \}\) are two bounded sequences of complex numbers, then
\[ \left(\sum_ {m = 1} ^ {\infty} \frac {a _ {m}}{m ^ {s}}\right) \left(\sum_ {k = 1} ^ {\infty} \frac {b _ {k}}{k ^ {s}}\right) = \sum_ {n = 1} ^ {\infty} \frac {c _ {n}}{n ^ {s}} \quad \text { where } c _ {n} = \sum_ {m k = n} a _ {m} b _ {k}. \]
The above series converge absolutely when \(\operatorname { R e } ( s ) > 1\)
- Prove as a consequence that one has
\[ (\zeta (s)) ^ {2} = \sum_ {n = 1} ^ {\infty} \frac {d (n)}{n ^ {s}} \quad \text { and } \quad \zeta (s) \zeta (s - a) = \sum_ {n = 1} ^ {\infty} \frac {\sigma_ {a} (n)}{n ^ {s}} \]
for \(\operatorname { R e } ( s ) > 1\) and \(\operatorname { R e } ( s - a ) > 1\) , respectively. Here \(d ( n )\) equals the number of divisors of \(n _ { \mathrm { : } }\) , and \(\sigma _ { a } ( n )\) is the sum of the \(a ^ { \mathrm { t h } }\) powers of divisors of \(n\) . In particular, one has \(\sigma _ { 0 } ( n ) = d ( n )\)
Solution
Part (a).
1. For \(\operatorname{Re}(s) > 1\), both series \(\sum_m \frac{a_m}{m^s}\) and \(\sum_k \frac{b_k}{k^s}\) converge absolutely.
\(|a_m|, |b_k|\) are bounded, and \(\sum \frac{1}{n^{\operatorname{Re} s}}\) converges for \(\operatorname{Re} s > 1\).
2. Hence their product is the Cauchy product:\[\left(\sum_m \frac{a_m}{m^s}\right)\left(\sum_k \frac{b_k}{k^s}\right) = \sum_{m,k} \frac{a_m b_k}{(mk)^s} = \sum_{n=1}^{\infty} \frac{c_n}{n^s},\]where \(c_n = \sum_{mk = n} a_m b_k\).
absolute convergence justifies rearranging the double sum by grouping terms with \(mk = n\).
Part (b).
1. \(\zeta(s)^2 = \left(\sum_m \frac{1}{m^s}\right)\left(\sum_k \frac{1}{k^s}\right) = \sum_n \frac{d(n)}{n^s}\).
by part (a) with \(a_m = b_k = 1\), we get \(c_n = \sum_{mk = n} 1 = d(n)\) (the number of divisors of \(n\)).
2. \(\zeta(s)\zeta(s-a) = \left(\sum_m \frac{1}{m^s}\right)\left(\sum_k \frac{k^a}{k^s}\right) = \sum_n \frac{\sigma_a(n)}{n^s}\).
by part (a) with \(a_m = 1\) and \(b_k = k^a\), we get \(c_n = \sum_{mk = n} k^a = \sigma_a(n)\) (the sum of the \(a\)-th powers of the divisors of \(n\)).
3. Q.E.D.
2 (a) and <1>1–<1>2 (b).