3.4. The limit and series versions of the fundamental theorem;
the prime number theorem

The limit and series versions

We have already obtained the integral version of the fundamental theorem for Dirichlet series, in the alternative forms 3.2.5 and 3.3.4. We will now deduce the limit version, stating (in the same notation) that

$$\frac{A(x)}{x} \to \alpha \quad \mbox{as } x \to \infty$$

and at the same time the series version, stating that

$$\sum_{n=1}^\infty \frac{a(n) - \alpha }{n}$$

converges (to $\alpha_0 - \gamma \alpha$). We can then read off what these statements say for special choices of Dirichlet series. In particular, when $a(n) = \Lambda (n)$, the limit version will give the desired fact that $\lim_{x \to \infty }\psi (x)/x = 1$. So the prime number theorem is obtained as just one example of a family of statements of this kind (though certainly the most important one).

The connecting link is provided by the following result, which is a Tauberian theorem of a much more elementary type than the Ingham-Newman theorem.

Proposition 3.4.1. Suppose that A is a real-valued function on $[1,\infty )$ such that for some $\alpha$,

$$\int_1^\infty \frac{A(x) - \alpha x}{x^2}\; dx$$

is convergent. Suppose also that

either	(i)	A is increasing and non-negative,
or	(ii)	there is another function B such that B and $B-A$ are increasing
		and non-negative and for some $\beta$, $\; \int_1^\infty [B(x) - \beta x]/x^2\; dx$ converges.

Then    $${ \frac{A(x)}{x} \to \alpha }$$  as  $x \to \infty$.

Proof. Firstly, statement (ii) follows at once from (i) because we then have

$$\frac{B(x)}{x} \to \beta \quad \mbox{and} \quad \frac{B(x) - A(x)}{x} \to \beta - \alpha \quad \mbox{as } x \to \infty,$$

hence $A(x)/x \to \alpha$.

We prove (i) first for the case $\alpha \neq 0$. It is enough to prove it when $\alpha = 1$; other cases are then derived by considering $A(x)/\alpha$. Let $0 < \delta < \frac{1}{2}$. Convergence of the stated integral implies that for any $\delta_1 > 0$, there exists $R$ such that

$$\left\vert \int_{x_0}^{x_1} \frac{A(x)- x}{x^2}\; dx \right\vert < \delta_1$$

whenever $x_1 > x_0 \geq R$. Suppose that for some $x_0 > R$ we have $A(x_0) > (1+ \delta )x_0$. Since $A(x)$ is increasing, $A(x) > (1 + \delta )x_0$ for all $x \geq x_0$. Let $x_1 = (1+ \delta )x_0$. Then

$$\begin{eqnarray*} \int_{x_0}^{x_1} \frac{A(x) - x}{x^2}\; dx & > & (1+ \delta )... ...ht) - \log \frac{x_1}{x_0} \\ & = & \delta - \log (1+ \delta ). \end{eqnarray*}$$

Now take $\delta_1 = \delta - \log (1+ \delta )$. Then $\delta_1 > 0$ (in fact, from the series for $\log (1+\delta )$, one has $\delta_1 \geq \frac{1}{3} \delta ^2$). So if $R$ corresponds to $\delta_1$ as above, then $A(x) \leq (1 + \delta )x$ for all $x > R$.

Similarly, if $A(x_0) < (1 - \delta )x_0$ for some $x_0 \geq 2R$, we take $x_2 = (1- \delta )x_0$ (note that $x_2 \geq R$), and we find that

$$\int_{x_2}^{x_0} \frac{A(x) - x}{x^2}\; dx \leq -\delta_2 ,$$

where  $\delta_2 = -\log (1- \delta ) - \delta \; (> \frac{1}{2}\delta^2)$. If $R$ corresponds to $\delta_2$ as before, we deduce that $A(x) \geq (1- \delta )x$ for all $x \geq 2R$.

Finally, the case $\alpha = 0$ is easy. Given $\delta$, there exists $R$ such that

$$\int_{x_0}^\infty \frac{A(x)}{x^2}\; dx < \delta$$

whenever $x_0 \geq R$. Suppose that  $A(x_0) \geq \delta x_0$  for some $x_0$. Then

$$\int_{x_0}^\infty \frac{A(x)}{x^2}\; dx \geq \delta x_0 \int_{x_0}^\infty \frac{1}{x^2}\; dx = \delta .$$

So we must have $A(x) \leq \delta x$ for $x \geq R$. $\Box$

We can now derive the limit and series statements, as promised. Actually, the series statement itself can be expressed in either of two equivalent forms, both of which have their uses. Conditions (FT1), (FT2), (FT3), (FT$3'$) are as stated in theorems 3.2.5 and 3.3.4.

Theorem 3.4.2. Suppose that $a(n)$ (real or complex) and $f(s)$ satisfy conditions (FT1),(FT2),(FT3), and that

either	(FT4)	$a(n) \geq 0$ for all $n$,
or	(FT5)	there exist $b(n)$, $g(s)$ satisfying conditions (FT1), (FT2), (FT3),
		with $\vert a(n)\vert \leq b(n)$ for all $n$.

Then:

(i)	$${ \frac{A(x)}{x} \to \alpha }$$  as $x \to \infty$,
(ii)	$${\sum_{n \leq x} \frac{a(n)}{n} - \alpha \log x \to \alpha_0}$$  as $x \to \infty$.
(iii)	$${ \sum_{n=1}^\infty \frac{a(n) - \alpha }{n} = \alpha_0 - \gamma \alpha }$$, where $\gamma$ is Euler's constant.

Condition (FT3) can be replaced by (FT$3'$) each time.

Proof. By 3.2.5 and 3.3.4, the integral hypothesis of 3.4.1 is satisfied. Under condition (FT4),  $A(x)$ is increasing, so  $\lim_{x \to \infty } [A(x)/x] = \alpha$. Under condition (FT5), if $a(n)$ is real, then $B(x)$ and  $B(x) - A(x)$  are increasing, so the second version of 3.4.1 applies to give the same conclusion. If $a(n)$ is complex, we obtain the same result by applying this to Re $A(x)$ and Im $A(x)$ separately.

By Abel's summation formula (1.3.6), we now have in either case

$$\begin{eqnarray*} \sum_{n\leq x} \frac{a(n)}{n} - \alpha \log x & = & \frac{A(x... ...(\alpha _0 - \alpha ) = \alpha_0 \quad \mbox{as } x \to \infty . \end{eqnarray*}$$

Hence also

$$\begin{eqnarray*} \sum_{n \leq x} \frac{a(n) - \alpha }{n} & = & \sum_{n \leq x... ...\alpha_0 - \gamma \alpha \quad \mbox{as } x \to \infty . \; \Box \end{eqnarray*}$$

Remark 1. For the purposes of the limit statement (i), there is no need to know $\alpha_0$ explicitly.

Remark 2. In the case when $f(s)$ has no pole at 1 (so that $\alpha = 0$), $\alpha_0$ is simply $f(1)$, and the series statement (in either form) becomes $\sum_{n=1}^\infty a(n)/n = f(1)$. In other words, the original Dirichlet series is also valid at $s = 1$.

Remark 3. In the special cases considered below, we will check both the alternative conditions (FT3) and (FT$3'$), so that either form of the integral theorem is sufficient for the conclusion.

The prime number theorem

We have now reached our original objective.

Theorem 3.4.3 (the Prime Number Theorem). Let $\psi$ be Chebyshev's function. Then

$$\frac{\psi (x)}{x} \to 1 \quad \mbox{as } x \to \infty ,$$

$$\pi (x) \sim \mbox{li}(x) \quad \mbox{as } x \to \infty .$$

Proof. We show that the first statement is a case of 3.4.2. As we saw in 1.6.2 and 2.4.5, the second statement then follows. Let $f(s) = - \zeta '(s)/\zeta (s)$. Then

$$f(s) = \sum_{n=1}^\infty \frac{\Lambda (n)}{n^s} \quad \mbox{for Re}\; s > 1$$

and

$$f(s) = \frac{1}{s-1} -\gamma + (s-1)h(s),$$

with $h(s)$ differentiable at 1. Of course, $\Lambda (n) \geq 0$ for all $n$, and $\sum_{n\leq x} \Lambda (n) = \psi (x)$. To verify (FT3), note that by 3.1.8 and 3.1.14, $\vert f(\sigma \pm it)\vert \leq P(t)$  when $\sigma \geq 1, \; t \geq 2$, where  $P(t) = 2(\log t +5)^9$. The integral $\int_1^\infty P(t)/t^2 \; dt$  is convergent, as required. Alternatively, Chebyshev's upper estimate $\psi (x) \leq 2x$ (2.4.6) shows that (FT$3'$) is satisfied. $\Box$

The reader may be tempted to pause for celebration!

As we saw in section 1.5, it follows that equally

$$\pi (x) \sim \frac{x}{\log x} \qquad \mbox{and} \qquad \pi (x) \sim \frac{x}{\log x -1}.$$

The prime number theorem is often quoted in one of these forms (especially the first one).

An obvious piece of unfinished business is the rate of convergence. We return to this question in chapter 5, where we will show that  $\vert\pi (x) - \mbox{li}(x)\vert$  is ultimately small compared with  $x/(\log x)^k$  for every positive $k$. Once this is known, it is clear that  li$(x)$  is a better approximation than $x/(\log x -1)$, which in turn is better than $x/\log x$.

Needless to say, the prime number theorem has interesting consequences and applications. We describe some in section 3.5.

At the same time, we have of course obtained integral and series statements relating to the Dirichlet series $\sum_{n=1}^\infty \Lambda(n)/n^s$. We record them here.

Proposition 3.4.4. For the Chebyshev function $\psi$ and the von Mangoldt function $\Lambda (n)$, we have:

$$\int_1^\infty \frac{\psi (x) - x}{x^2}\; dx = - \gamma -1,$$

$$\sum_{n\leq x} \frac{\Lambda (n)}{n} - \log x \to -\gamma \quad \mbox{as } x \to \infty .$$

$$\sum_{n=1}^\infty \frac{\Lambda (n) -1}{n} = -2 \gamma. \; \Box$$

Further special cases of the basic theorem

The second most important case of the basic theorem 3.4.2 is found by applying it to $1/\zeta (s)$. This time we need condition (FT5) instead of (FT4).

Theorem 3.4.5. Let $\mu (n)$ be the Möbius function and $M(x) = \sum_{n\leq x}\mu (n)$. Then

$$\int_1^\infty \frac{M(x)}{x^2}\; dx = 0,$$

$$\frac{M(x)}{x} \to 0 \quad \mbox{as } x \to \infty ,$$

$$\sum_{n=1}^\infty \frac{\mu (n)}{n} = 0.$$

Proof. Let  $f(s) = 1/\zeta (s)$  for $s \neq 1$, and $f(1) = 0$. Then

$$f(s) = \sum_{n=1}^\infty \frac{\mu (n)}{n^s} \quad \mbox{for Re}\; s > 1$$

and  $f(s) = 0 + (s-1)h(s)$, with $h(s)$ differentiable at 1. By 3.1.14, condition (FT3) is satisfied, with  $P(t) = 4(\log t +5)^7$. The alternative condition (FT$3'$) is trivial, since obviously $\vert M(x)\vert \leq x$. Also, since $\vert\mu (n)\vert \leq 1$ for all $n$, we can take $b(n) = 1$ and $g(s) = \zeta (s)$ in (FT5): these functions clearly satisfy the required conditions. The statements follow. $\Box$

Similar statements apply to the Liouville function:

Proposition 3.4.6. Let $\lambda (n)$ be the Liouville function, and let $S_\lambda (x) = \sum_{n \leq x} \lambda (n)$. Then

$$\int_1^\infty \frac{S_\lambda (x)}{x^2}\; dx = \lim_{x \to \i... ...\lambda (x)}{x} = \sum_{n=1}^\infty \frac{\lambda (n)}{n} = 0.$$

Proof. Let  $f(s) = \zeta (2s)/\zeta (s)$  for $s \neq 1$ and $f(1) = 0$. Then

$$f(s) = \sum_{n=1}^\infty \frac{\lambda (n)}{n^s} \quad \mbox{for Re}\; s > 1$$

and $f$ is holomorphic, with value 0, at 1. For condition (FT3), note that $\vert\zeta (2s)\vert \leq \zeta (2)$ when $\sigma \geq 1$. Again, (FT$3'$) is immediate. The statements follow, as before. $\Box$

Because $1/\zeta (s)$ has no pole at 1, we can extend these results to the whole line Re $s = 1$, as follows. We confine ourselves to the Möbius function (the Liouville function is similar), and to stating the series version, which amounts to saying that the original Dirichlet series converges on this line.

Proposition 3.4.7. For all non-zero, real t, we have

$$\sum_{n=1}^\infty \frac{\mu (n)}{n^{1+it}} = \frac{1}{\zeta (1+it)} .$$

Proof. Fix $t$. Let $a(n) = n^{-it} \mu (n)$ and  $f(s) = 1/\zeta (s+it)$. Then $f(s)$ is defined and differentiable on the whole line Re $s = 1$, and  $\sum_{n=1}^\infty [a(n)/n^s] = f(s)$  for Re $s > 1$. As for any holomorphic function, we have  $f(s) = f(1) + (s-1)h(s)$, where $h(s)$ is differentiable at 1, so (in the notation of 3.4.2) $\alpha = 0$ and $\alpha_0 = f(1)$. Also, if  $\vert u\vert \geq \vert t\vert + 2$, then

$$\vert f(\sigma + iu)\vert \leq 4 [\log(\vert t\vert + \vert u\vert + 5)]^7,$$

so condition (FT3) is satisfied; again, (FT$3'$) is trivial. As before, we can take $b(n) = 1$ in condition (FT5). The statement follows. $\Box$

In the same way, we can extend 3.4.4 by working with the function

$$f(s) = \zeta (s) + \frac{\zeta '(s)}{\zeta (s)}.$$

By adding the two series, one sees that this function has no pole at 1. With $t$ fixed, take

$$f_t(s) = f(s+it) = \sum_{n=1}^\infty \frac{a(n)}{n^s},$$

where  $a(n) = n^{-it}[1 - \Lambda (n)]$. The dominating sequence is $b(n) = 1+ \Lambda (n)$. We obtain:

Proposition 3.4.8. Let $f(s)$ be as just stated. For all non-zero, real t, we have

$$\sum_{n=1}^\infty \frac{1 - \Lambda (n)}{n^{1+it}} = f(1+it). \; \Box$$

Corollary 3.4.9. For $t \neq 0$, we have

$$\sum_{n=1}^N \frac{\Lambda (n)}{n^{1+it}} = - \frac{\zeta'(1+it)}{\zeta (1+it)} - \frac{1}{it}N^{-it} + r_N,$$

where  $r_N \to 0$ as $N \to \infty$.

Proof. Recall that by 3.1.3,

$$\sum_{n=1}^N \frac{1}{n^{1+it}} = \zeta (1+it) - \frac {1}{it} N^{-it} + q_N,$$

where $q_N \to 0$ as $N \to \infty$. The result follows by adding the sum of $N$ terms in 3.4.8. $\Box$

Hence the partial sums of $\sum \Lambda (n)/n^{1+it}$ eventually describe circles around the point $- (\zeta'/\zeta )(1+it)$ in the same way that those of $\sum 1/n^{1+it}$ do so around $\zeta (1+it)$.

In particular, these partial sums are bounded. Recall that for Re $s > 1$, we have $\log \zeta (s) = \sum_{n=1}^\infty c(n)/n^s$, where $c(n) = \Lambda (n)/\log n$. Since

$$\frac{c(n)}{n^s} = \frac{\Lambda (n)}{n^s} \; \frac{1}{\log n},$$

discrete Abel summation (in the form 1.3.4) shows that $\sum_{n=1}^\infty c(n)/n^s$ converges for $s \neq 1$ with Re $s = 1$. A routine argument shows that the sum of the series is continuous onto this line, from which it follows that it still equals $\log \zeta (s)$ there. It follows easily that the Euler product converges to $\zeta (s)$ for such $s$. (For the details, see e.g. [Titch], section 3.14.)

Exercises

1. Prove that  $\psi (x) - \theta (x) \sim x^{1/2}$ as $x \to \infty$.

2. Show that

$$\sum_{n \leq x} \frac{\Lambda (n) +1}{n} = 2 \log x + q(x),$$

where  $q(x) \to 0$  as $x \to \infty$.

3. Let $M_2(x) = \sum_{n \leq x}\vert\mu (n)\vert$. Use the theorems of the present section to re-prove 2.5.5 in the form

$$\frac{M_2(x)}{x} \to \frac{1}{\zeta (2)} = \frac{6}{\pi ^2} \quad \mbox{as } x \to \infty .$$

4. Let $a(n) = \mu (n)\log n$. What function has Dirichlet series $\sum_{n=1}^\infty a(n)/n^s$? Show that

$$\int_1^\infty \frac{A(x)}{x^2}\; dx = -1.$$

5. Prove that  $\sum_{p \in P} 1/p^s$  is convergent for $s \neq 1$ with Re $s = 1$

6. Let $E_k$ be the set of integers $n$ such that  $2^{k-1} < n \leq 2^k$. Define $a(n)$ to be 1 if $n$ is in $\bigcup _{k=1}^\infty E_{2k}$, and 0 otherwise. Let  $f(s) = \sum_{n=1}^\infty a(n)/n^s$  for Re $s > 1$. By applying Euler's summation formula to  $\sum_{n\in E_{2k}} 1/n^s$, show that $f(s)$ can be expressed as $g(s) + h(s)$, where

$$g(s) = \frac{1}{(s-1)(2^{s-1} +1)}$$

and $h(s)$ is holomorphic for Re $s > 0$ (so that $f(s)$ has a simple pole at 1). Show however that $A(N)/N$ does not tend to a limit as $N \to \infty$, by considering $N = 2^{2k}$ and $N = 2^{2k+1}$. Which of the hypotheses of the fundamental theorems is not satisfied?

7. Show by the following steps that the prime number theorem (if proved another way) implies that $\zeta (1 + it) \neq 0$ for all $t \neq 0$. Suppose that $\zeta (s)$ has a zero of order $m$ at $s_0 = 1 + it_0$. Let

$$f(s) = s \int_1^\infty \frac{\psi (x) - x}{x^{s+1}}\; dx.$$

Show that $(s - s_0)f(s) \to m$ as $s \to s_0$. However, assuming that $\psi (x) \sim x$, show that $(\sigma -1)\vert f(\sigma + it_0)\vert$ tends to 0 as $\sigma \to 1^+$, hence obtaining a contradiction.