Muhammad Haris Rao
For an arithmetical function $f : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}$, define $\mathfrak{D}_f (s)$ for $s \in \mathbb{C}$ to be the series \begin{align*} \mathfrak{D}_f \left( s \right) &= \sum_{n = 1}^\infty \frac{f(n)}{n^s} \end{align*} Of course, we will often need to be careful about convergence issues. By comparision to the series representation of the Riemann zeta function, $\mathfrak{D}_f$ converges to an analytic function for $\mathfrak{Re}(s) > 1$ when $f$ is bounded. It turns out that the region where the series converges is always a half plane to the right of some vertical line.
Lemma: Suppose that $f : \mathbb{Z}_{>0} \longrightarrow \mathbb{C}$ is an arithmetical function and $s_0 \in \mathbb{C}$ is a point in the complex plane are such the partial sums of the associated Dirichlet series at the point are bounded by $M$. That is, for all $N \in \mathbb{Z}_{>0}$, \begin{align*} \left| \sum_{n = 1}^N \frac{f(n)}{n^s} \right| \le M \end{align*} Then, it holds for all positive integers $a \le b$ that \begin{align*} \left| \sum_{n = 1}^b \frac{f(n)}{n^s} - \sum_{n = 1}^a \frac{f(n)}{n^s} \right| < \frac{2M}{a^{\sigma - \sigma_0}} \left( 1 + \sqrt{1 + \frac{(\tau - \tau_0)^2}{(\sigma - \sigma_0)^2} } \right) \end{align*} where $s = \sigma + i \tau$ and $s_0 = \sigma_0 + i \tau_0$.
Proof. The Abel summation formula yields \begin{align*} \sum_{a < n \le b} \frac{f(n)}{n^s} = \sum_{a < n \le b} \frac{1}{n^{s - s_0}} \frac{f(n)}{n^{s_0}} = \frac{1}{b^{s - s_0}} \sum_{n = 1}^b \frac{f(n)}{n^{s_0}} - \frac{1}{a^{s - s_0}} \sum_{n = 1}^a \frac{f(n)}{n^{s_0}} - \int_a^b \frac{s_0 - s}{t^{s - s_0 + 1}} \sum_{n \le t} \frac{f(n)}{n^s} \, dt \end{align*} Hence, taking absolute value gives \begin{align*} \left| \sum_{a < n \le b} \frac{f(n)}{n^s} \right| &\le \frac{1}{b^{\sigma - \sigma_0}} \left| \sum_{n = 1}^b \frac{f(n)}{n^{s_0}} \right| + \frac{1}{a^{\sigma - \sigma_0}} \left| \sum_{n = 1}^a \frac{f(n)}{n^{s_0}} \right| + | s - s_0 | \int_a^b \frac{1}{t^{\sigma - \sigma_0 + 1}} \left| \sum_{n \le t} \frac{f(n)}{n^s} \right| \, dt \\ &\le M \left( \frac{1}{b^{\sigma - \sigma_0}} + \frac{1}{a^{\sigma - \sigma_0}} + |s - s_0| \int_a^b \frac{1}{t^{\sigma - \sigma_0 + 1}} \, dt \right) \\ &\le M \left( \frac{2}{a^{\sigma - \sigma_0}} + |s - s_0| \left[ \frac{t^{\sigma_0 - \sigma}}{\sigma_0 - \sigma} \right]_a^b \right) \\ &= M \left( \frac{2}{a^{\sigma - \sigma_0}} + \frac{|s - s_0|}{\sigma - \sigma_0} \left( \frac{1}{a^{\sigma - \sigma_0}} - \frac{1}{b^{\sigma - \sigma_0}} \right) \right) \\ &< \frac{2M}{a^{\sigma - \sigma_0}} \left( 1 + \sqrt{1 + \frac{(\tau - \tau_0)^2}{(\sigma - \sigma_0)^2} } \right) \end{align*} $\space$ $\blacksquare$
This bound allows us to prove the following:
Theorem: Let $f$ be an arithmetical function, $s_0 \in \mathbb{C}$ and $M \ge 0$ a bound for the partial sums of the associated series at $s_0$ as in the statemetnt of the previous lemma. Then $\mathfrak{D}_f(s)$ converges uniformly on every compact subset to the right of $s_0$. Consequently, it defines a holomorphic function on this half plane with derivatives \begin{align*} \frac{d^k}{ds^k} \mathfrak{D}_f(s) &= \sum_{n = 1}^\infty \frac{(-1)^k f(n) \log^k{n}}{n^s} \end{align*}
Proof. Let $K \subseteq \mathbb{C}$ be compact with $\mathfrak{Re}(s) > \sigma_0$ for all $s \in K$. By compactness, we have \begin{align*} A &= \text{max} \{ |\tau - \tau_0| \mid \tau = \mathfrak{Im}(s), \text{ for $s \in K$} \} \\ B &= \text{min} \{ |\sigma - \sigma_0| \mid \sigma = \mathfrak{Re}(s), \text{ for $s \in K$} \} \end{align*} By compactness, it must be $B > 0$. Now set $C = A / B$. Then for any integers $a \le b$ we have for all $s \in K$ that \begin{align*} \left| \sum_{n = 1}^b \frac{f(n)}{n^s} - \sum_{n = 1}^a \frac{f(n)}{n^s} \right| < \frac{2M}{a^{\sigma - \sigma_0}} \left( 1 + \sqrt{1 + \frac{(\tau - \tau_0)^2}{(\sigma - \sigma_0)^2} } \right) \le \frac{2M}{a^B} \left( 1 + \sqrt{1 + \frac{A^2}{B^2} } \right) = \frac{2M\left( 1 + \sqrt{1 + C^2} \right)}{a^B} \end{align*} Given $\varepsilon > 0$, choose an integer $N \ge \left( \frac{2M}{\varepsilon} \left( 1 + \sqrt{1 + C^2} \right) \right)^{1/B}$. Then we have for every $N \le a \le b$ and every $s \in K$ that \begin{align*} \left| \sum_{n = 1}^b \frac{f(n)}{n^s} - \sum_{n = 1}^a \frac{f(n)}{n^s} \right| < \frac{2M\left( 1 + \sqrt{1 + C^2} \right)}{a^B} \le \frac{2M\left( 1 + \sqrt{1 + C^2}\right) }{N^B} \le \frac{2M\left( 1 + \sqrt{1 + C^2}\right)}{\left( \left( \frac{2M}{\varepsilon} \left( 1 + \sqrt{1 + C^2} \right) \right)^{1/B} \right)^B} = \varepsilon \end{align*} This proves that the sequence of partial sums is uniformly Cauchy on $K$. Hence, it converges uniformly on this arbitrary compact set, which is what was claimed. $\blacksquare$
This result implies that the interir of the region on which a Dirichlet series converges is an open right half plane. Therefore, we define the abscissa of convergence for a given arithmetical function is \begin{align*} \sigma_f &= \text{inf} \left\{ \mathfrak{Re}(s) \mid \sum_{n = 1}^\infty \frac{f(n)}{n^s} \text{ converges } \right\} \\ &= \text{inf} \left\{ \mathfrak{Re}(s) \mid \text{ the sequence of partial sums of $\sum_{n = 1}^\infty \frac{f(n)}{n^s}$ are bounded } \right\} \end{align*} where teh second equality is by the previous theorem. The Dirichlet series converges and defines a holomrphic function on the half plane to the right side of $\sigma_f$, and diverges everywhere on the left. We define also the abscissa of absolute convergence \begin{align*} \sigma_f^a = \text{inf} \left\{ \mathfrak{Re}(s) \mid \sum_{n = 1}^\infty \frac{f(n)}{n^s} \text{ converges absolutely } \right\} \\ \end{align*} We have the following fact about the abscissa of absolute convergence:
Proposition: For any arithmetic function $f$, $\sigma_f^a = \sigma_{|f|}$.
Proof. If $f \ge 0$, then if we were to have $\sigma_f < \sigma_f^a$, there would be a real number $\sigma \in (\sigma_f, \sigma_f^a)$ where the Dirichlet series of $f$ would converge but not absolutely. This would be absurd since all the summands are positive. Hence, for any arithmetic function $f$, $\sigma_{|f|} = \sigma_{|f|}^a$. Therefore, \begin{align*} \sigma_f^a &= \text{inf} \left\{ \mathfrak{Re}(s) \mid \sum_{n = 1}^\infty \left| \frac{f(n)}{n^s} \right| < \infty \right\} = \text{inf} \left\{ \mathfrak{Re}(s) \mid \sum_{n = 1}^\infty \left| \frac{|f(n)|}{n^s} \right| < \infty \right\} = \sigma_{|f|}^a = \sigma_{|f|} \end{align*} which is the desired equality. $\blacksquare$
We have the following result about the existence of abscissa of convergence.
Proposition: Let $f$ be an arithmetical function, and suppose $r \in \mathbb{R}$ is such that $f(n) = \mathcal{O}(n^{r + \delta})$ for every $\delta > 0$. Then the Dirichlet series for $f$ converges absolutely on the right of $1 + r$. Consequently, $\sigma_f^a \le 1 + r$.
Proof. Let $s \in \mathbb{C}$ be such that $\mathfrak{Re}(s) > 1 + r$. Let $0 < \varepsilon < \mathfrak{Re}(s) - (1 + r)$ so that $\mathfrak{Re}(s) > 1 + r + \varepsilon$. Taking $\delta = \varepsilon / 2$, there is $M \ge 0$ such that $|f(n)| \le M n^{r + \delta}$. Therefore, if $N \in \mathbb{Z}_{> 0}$, then $$ \sum_{n = 1}^N \left| \frac{f(n)}{n^s} \right| \le \sum_{n = 1}^N \frac{M n^{1 + \delta}}{n^{\mathfrak{Re}(s)}} \le \sum_{n = 1}^N \frac{M n^{r + \varepsilon/2}}{n^{1 + r + \varepsilon}} \le M \sum_{n = 1}^\infty \frac{1}{n^{1 + \varepsilon/2}} < \infty $$ Then taking $N \to \infty$, the sum on the left must converge since the summands are non-negative and the partial sums have now been shown to be bounded. Therefore, the Dirichlet series of $f$ converges absolutely at any $s \in \mathbb{C}$ with $\mathfrak{Re}(s) > 1 + r$. So then $\sigma_f^a \le 1 + r$. $\blacksquare$
This yields the following inequalities on the abscissa of convergence: \begin{align*} \sigma_f \le \sigma_f^a \le 1 + \text{inf}\left\{ r \in \mathbb{R} \mid f(n) = \mathcal{O}(n^{r + \delta}) \text{ for every $\delta > 0$} \right\} \end{align*}
Recall that the convolution of $f, g : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}$ is defined as \begin{align*} f * g : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}, n \longmapsto \sum_{d \mid n} f\left( d \right) g \left( \frac{n}{d} \right) \end{align*} Given that $f(n) = \mathcal{O}(n^{r_1})$ and $g(n) = \mathcal{O}(n^{r_2})$, we would like to find the growth order of $f * g$. For this, we define the arithmetic function $d(n)$ to be the number of divisors of $n \in \mathbb{Z}_{> 0}$. We state without proof that
Theorem: For every $\delta > 0$, $d(n) = \mathcal{O}(n^\delta)$.
From this, we can prove
Theorem: Suppose that $f, g : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}$ with $f(n), g(n) = \mathcal{O}(n^{r + \delta})$ for every $\delta> 0$. Then also $(f * g)(n) = \mathcal{O}(n^{r + \delta})$ for every $\delta > 0$
Proof. Let $\delta > 0$ be arbitrary so that there is $M_1, M_2, M_3 \ge 0$ with $f(n) \le M n^{r + \delta/2}$, $g(n) \le M_2 n^{r + \delta/2}$ and $d(n) \le M_3 n^{\delta/2}$. Then it follows that \begin{align*} \left| \left( f * g \right)(n) \right| &= \left| \sum_{d \mid n} f\left( d \right) g \left( \frac{n}{d} \right) \right| \\ &\le \sum_{d \mid n} \left| f\left( d \right) \right| \left| g \left( \frac{n}{d} \right) \right| \\ &\le \sum_{d \mid n} \left( M_1 d^{r + \delta / 2} \right) \left( M_1 \left( \frac{n^{r + \delta / 2}}{d^{r + \delta/2}} \right) \right) \\ &= M_1 M_2 n^{r + \delta/2} \sum_{d \mid n} 1 \\ &= M_1 M_2 n^{r + \delta/2} d(n) \\ &\le M_1 M_2 M_3 n^{r + \delta} \end{align*} which is $\mathcal{O}(n^{r + \delta})$ growth. This applies to arbitrary $\delta > 0$. $\blacksquare$
Hence, we can say that
Corollary: The set of arithmetic functions with growth order $\mathcal{O}(n^{r + \delta})$ for every $\delta > 0$ is closed under pointwise addition and Dirichlet convolution. Hence, they form a unital commutative ring with these operations.
The important fact about Dirichlet convlution is the following:
Theorem: Let $f, g : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}$ be arithmetic functions with $f, g = \mathcal{O}(n^{r + \delta})$ for every $\delta > 0$. Then, we have \begin{align*} \left( \sum_{n = 1}^\infty \frac{f(n)}{n^s} \right) \left( \sum_{n = 1}^\infty \frac{g(n)}{n^s} \right) &= \sum_{n = 1}^\infty \frac{(f * g)(n)}{n^s} \end{align*} for all $s \in \mathbb{C}$ with $\mathfrak{Re}(s) > 1 + r$. The convergence of each series is in the absolute sense.
The fact that all the sums converge absolutely on the specified domain follows from the previous corollary. Observe that \begin{align*} \left| \left( \sum_{k = 1}^n \frac{f(k)}{k^s} \right) \left( \sum_{k = 1}^n \frac{g(k)}{k^s} \right) - \sum_{k = 1}^n \frac{(f*g)(k)}{k^s} \right| &= \left| \sum_{k = n + 1}^{n^2} \left( \sum_{d \mid k, d \le n, k/d \le n} f \left( d \right) g \left( \frac{k}{d} \right) \right) \frac{1}{k^s} \right| \\ &\le \sum_{k = n + 1}^{n^2} \left( \sum_{d \mid k} \left| f \left( d \right) g \left( \frac{k}{d} \right) \right| \right) \frac{1}{k^s} \\ &\le \sum_{k = n + 1}^\infty \frac{(|f| * |g|)(k)}{k^\sigma} \end{align*} where $\sigma = \mathfrak{Re}(s) > \sigma_f^a$. Thus, the above expression vanishes as $n \to \infty$. By the inequality \begin{align*} \left| \left( \sum_{k = 1}^n \frac{f(k)}{k^s} \right) \left( \sum_{k = 1}^n \frac{f(k)}{k^s} \right) - \sum_{k = 1}^\infty \frac{(f*g)(k)}{k^s} \right| &\le \left| \left( \sum_{k = 1}^n \frac{f(k)}{k^s} \right) \left( \sum_{k = 1}^n \frac{f(k)}{k^s} \right) - \sum_{k = 1}^n \frac{(f*g)(k)}{k^s} \right| + \left| \sum_{k = 1}^n \frac{(f*g)(k)}{k^s} - \sum_{k = 1}^\infty \frac{(f*g)(k)}{k^s} \right| \\ &\le \sum_{k = n + 1}^\infty \frac{(|f| * |g|)(k)}{k^\sigma} + \left| \sum_{k = 1}^n \frac{(f*g)(k)}{k^s} - \sum_{k = 1}^\infty \frac{(f*g)(k)}{k^s} \right| \\ \end{align*} everything vanishes as $n \to \infty$ so that \begin{align*} \left( \sum_{n = 1}^\infty \frac{f(n)}{n^s} \right) \left( \sum_{n = 1}^\infty \frac{f(n)}{n^s} \right) &= \sum_{n = 1}^\infty \frac{(f*g)(n)}{n^s} \end{align*} as was claimed. $\blacksquare$
Recall that an arithmetic function $f : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}$ is said to be multiplicative if $f(n m) = f(n) f(m)$ for any coprime positive integers $n, m$, and is said to be completely multiplicative if $f(nm) = f(n) f(m)$ for all positive integers $n, m$ regardless of coprimality. We have the following fact:
Theorem: Let $f : \mathbb{Z}_{> 0}$ be an arithmetic function. Then on the half plane of absolute convergence, its associated Dirichlet series factorises as \begin{align*} \sum_{n = 1}^\infty \frac{f(n)}{n^s} &= \prod_p \left( \sum_{n = 1}^\infty \frac{f\left( p^n \right)}{p^{ns}} \right) \end{align*} if $f$ is multiplicative and \begin{align*} \sum_{n = 1}^\infty \frac{f(n)}{n^s} &= \prod_p \frac{1}{1 - f(p) p^{-s}} \end{align*} if $f$ is completely mulitplicative.
Proof. Suppose $f$ is multiplicative and $\mathfrak{Re}(s) > \sigma_f^a$. Let $N, M \in \mathbb{Z}_{> 0}$ and let $S_{N_1, N_2}$ be the set of integers which can be expressed as a product of the primes less than or equal to $N$ each with mulitplicity at most $M$. It follows from the fundamental theroem of arithmetic that \begin{align*} \prod_{p \le N} \left( \sum_{n = 0}^{M} \frac{f\left( p^n \right)}{p^{ns}} \right) = \sum_{n \in S_{N, M}} \frac{f(n)}{n^s} \end{align*} Each of the sums in the finite product on the left will converge absolutely if we take $M \to \infty$ by the following bound \begin{align*} \sum_{n = 0}^{M} \left| \frac{f\left( p^n \right)}{p^{ns}} \right| \le \sum_{n = 1}^{p^M} \left| \frac{f(n)}{n^s} \right| \le \sum_{n = 1}^{\infty} \left| \frac{f(n)}{n^s} \right| < \infty \end{align*} Hence, we can take the limit as $M \to \infty$ and obtain \begin{align*} \prod_{p \le N} \left( \sum_{n = 0}^{\infty} \frac{f\left( p^n \right)}{p^{ns}} \right) = \lim_{M \to \infty} \sum_{n \in S_{N, M}} \frac{f(n)}{n^s} = \sum_{n \in T_N} \frac{f(n)}{n^s} \end{align*} where $T_N \subseteq \mathbb{Z}_{> 0}$ is the positive integers whose prime factors are bounded by $N$. The last limit is not too difficult to prove directly, but it can also be deduced by considering the sum as a Lebesgue integral, noting that the sequence of sets $\{ S_{N, M} \}_{M \ge 1}$ is increasing to $T_N$ and then applying Lebesgue's dominated convergence theorem. Similarly, the sets $\{ T_N \}_{N \ge 1}$ converge to $\mathbb{Z}_{> 0}$ so we can apply the same reasoning to get \begin{align*} \prod_p \left( \sum_{n = 0}^{\infty} \frac{f\left( p^n \right)}{p^{ns}} \right) &= \lim_{N \to \infty} \sum_{n \in T_N} \frac{f(n)}{n^s} = \sum_{n = 1}^\infty \frac{f(n)}{n^s} \end{align*} which is the desired equality. In the case where $f$ is completely multiplicative, we can consider the infinite sums as geometric series to get \begin{align*} \sum_{n = 1}^\infty \frac{f(n)}{n^s} = \prod_p \left( \sum_{n = 0}^{\infty} \left( \frac{f\left( p \right)}{p^s} \right)^n \right) = \prod_p \frac{1}{1 - f(p) p^{-s}} \end{align*} which is what we set out to prove. $\blacksquare$
A consequence of this result is that the Riemann Zeta function factorises as \begin{align*} \zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s} = \prod_p \frac{1}{1 - p^{-s}} \end{align*} for all $s \in \mathbb{C}$ with $\mathfrak{Re}(s) > 1$. Since the Möbius function is multiplicative, we have \begin{align*} \sum_{n = 1}^\infty \frac{\mu(n)}{n^s} &= \prod_p \left( \sum_{n = 1}^\infty \frac{\mu\left( p^n \right)}{p^{ns}} \right) = \prod_p \left( 1 - p^{-s} \right) \end{align*} since $\mu(p^n)$ is 1 for $n = 0$, $-1$ for $n = 1$, and 0 otherwise. Comparing this to the Euler product for $\zeta(s)$ gives another proof that the Dirichlet series for the Möbius function is the reciprocal of the Riemann zeta function.
The Euler totient function $\varphi(n)$ giving the number of integers at most $n$ which are coprime to $n$ is multiplicative. Furthermore, it is not hard to calculate \begin{align*} \varphi \left( p^n \right) = p^n - p^{n - 1} \end{align*} Therefore, \begin{align*} \sum_{n = 1}^\infty \frac{\varphi(n)}{n^s} &= \prod_p \left( \sum_{n = 0}^\infty \frac{\varphi \left( p^n \right)}{p^{ns}} \right) \\ &= \prod_p \left( 1 + \sum_{n = 1}^\infty \frac{p^n - p^{n - 1}}{p^{ns}} \right) \\ &= \prod_p \left( 1 + \sum_{n = 1}^\infty \frac{1}{p^{n(s - 1)}} - \frac{1}{p} \sum_{n = 0}^\infty \frac{1}{p^{n(s - 1)}} \right) \\ &= \prod_p \left( 1 + \frac{p^{1 - s}}{1 - p^{1 - s}} - \frac{1}{p} \frac{p^{1 - s}}{1 - p^{1 - s}} \right) \\ &= \prod_p \frac{1 - p^{-s}}{1 - p^{1 - s}} \\ &= \frac{\prod_p \frac{1}{1 - p^{-(s - 1)}}}{\prod_p \frac{1}{1 - p^{-s}}} \\ &= \frac{\zeta(s - 1)}{\zeta(s)} \end{align*} Because $\varphi(n) = \mathcal{O}(n)$, the above equality is valid for at least $\sigma > 2$. This tells us that \begin{align*} \sum_{n = 1}^\infty \frac{\varphi(n)}{n^s} &= \left( \sum_{n = 1}^\infty \frac{n}{n^s} \right) \left( \sum_{n = 1}^\infty \frac{\mu(n)}{n^s} \right) \end{align*} By the uniqueness theorem for Dirichlet series, this implies \begin{align*} \varphi(n) &= \left( \text{Id} * \mu \right)(n) = n \sum_{d \mid n} \frac{\mu(d)}{d} \end{align*}
Recall that for $r \in \mathbb{R}$, \begin{align*} \sigma_r \left( n \right) &= \sum_{d \mid n} d^r \end{align*} Thus, by the theorem linking products of Dirichlet series and Dirichlet convolution, \begin{align*} \sum_{n = 1}^\infty \frac{\sigma_r(n)}{n^s} &= \left( \sum_{n = 1}^\infty \frac{1}{n^s} \right) \left( \sum_{n = 1}^\infty \frac{1}{n^{s - r}} \right) = \prod_p \frac{1}{(1 - p^{-s}) (1 - p^{-(s - r)})} = \zeta(s) \zeta(s - r) \end{align*} valid for $\sigma > r$.
The Liouville funciton $\lambda(n)$ counts the whether the number of primes counted with mulitiplicity in the prime factorisation of $n$ is even or odd. In the first case, $\lambda(n) = 1$ and $\lambda(n) = -1$ in the other case. Clearly, $\lambda$ is completely multiplicative so that \begin{align*} \sum_{n = 1}^\infty \frac{\lambda(n)}{n^s} &= \prod_p \frac{1}{1 - \lambda(p) p^{-s}} =\prod_p \frac{1}{1 + p^{-s}} = \prod_p \frac{1 - p^{-s}}{1 - p^{-2s}} = \frac{\zeta(2s)}{\zeta(s)} \end{align*} Since $\lambda(n) = \mathcal{O}(1)$, this is valid for $\sigma > 1$.