Analytic Number Theory

Proof. Fix real $\sigma >0$. The function $x\mapsto x^{-\sigma }$ is positive and decreasing on $[1,\infty )$, so for every integer $n\ge 2$,

Summing the right-hand inequality over $2\le n\le N$ and adding the term $n=1$ gives

and summing the left-hand inequality over $1\le n\le N$ gives ${\sum }_{n=1}^N{n}^{-\sigma }\ge {\int }_1^{N+1}{x}^{-\sigma }dx$.

If $\sigma >1$ then $N^{1-\sigma }\to 0$, so the partial sums are bounded above by $1+1/(\sigma -1)$ and are increasing, hence convergent; and the lower bound tends to $1/(\sigma -1)$. Since $|n^{-s}|=n^{-\sigma }$, convergence of $\sum n^{-\sigma }$ is absolute convergence of $\sum n^{-s}$.

If $\sigma =1$ the lower bound is $\log (N+1)\to \infty$, and if $\sigma <1$ it is $((N+1{)}^{1-\sigma }-1)/(1-\sigma )\to \infty$. In both cases the series diverges.∎

Proof. On $\sigma \ge 1+\delta$ we have $|n^{-s}|=n^{-\sigma }\le n^{-1-\delta }$, and $\sum n^{-1-\delta }$ converges by the theorem Convergence of the zeta series. The Weierstrass $M$ -test therefore gives uniform convergence there. Each partial sum ${\sum }_{n\le N}{n}^{-s}={\sum }_{n\le N}{e}^{-s\log n}$ is entire, and a locally uniform limit of holomorphic functions is holomorphic with termwise-convergent derivatives, by Weierstrass' convergence theorem. Differentiating $e^{-s\log n}$ in $s$ gives $-\log n\cdot n^{-s}$.∎

Proof. For $\sigma >1$ the integral ${\int }_1^{\infty }{x}^{-s}dx$ converges and equals $1/(s-1)$. Splitting it over the intervals $[n,n+1]$,

which is legitimate because both series converge absolutely there. Now estimate a single ${\delta }_n$ for arbitrary $s$ with $\sigma >0$. Since ${\delta }_n(s)={\int }_n^{n+1}\left(n^{-s}-x^{-s}\right)dx$ and

we get $|n^{-s}-x^{-s}|\le |s|{\int }_n^{n+1}{u}^{-\sigma -1}du\le |s|n^{-\sigma -1}$ for $n\le x\le n+1$, hence

On a compact subset of $\{\sigma >0\}$ we have $|s|\le M$ and $\sigma \ge \delta >0$, so $|{\delta }_n(s)|\le M{n}^{-1-\delta }$, a convergent bound independent of $s$. By the $M$ -test the sum converges uniformly there, and each ${\delta }_n$ is holomorphic on $\sigma >0$, so the sum is too. It agrees with $\zeta (s)-1/(s-1)$ on $\sigma >1$, so it is the continuation.∎

Proof. By the theorem Continuation to $\sigma >0$, $\zeta (s)=1/(s-1)+h(s)$ with $h$ holomorphic on $\sigma >0$. A holomorphic $h$ plus a simple pole is a simple pole, and multiplying by $s-1$ gives $(s-1)\zeta (s)=1+(s-1)h(s)\to 1$. Finally $h(1)={\sum }_{n\ge 1}\left(\frac{1}{n}-\log \frac{n+1}{n}\right)={\lim }_N\left({\sum }_{n\le N}\frac{1}{n}-\log (N+1)\right)=\gamma$, the sum telescoping in its logarithms; this is precisely the definition of $\gamma$.∎

Proof. For $\sigma >1$ all series converge absolutely, so we may rearrange:

since subtracting $\eta$ from $\zeta$ leaves twice the even terms. Rearranging gives $\eta (s)=(1-2^{1-s})\zeta (s)$. That $\sum (-1{)}^{n-1}{n}^{-s}$ converges for $\sigma >0$ follows from summation by parts: the partial sums of $(-1{)}^{n-1}$ are bounded by $1$, and $n^{-s}$ has bounded variation tending to $0$ along $[1,\infty )$ when $\sigma >0$.∎

Proof. Proof sketch — assumes the theta transformation and Mellin inversion; not elementary. Start from $\Gamma (s/2){\pi }^{-s/2}{n}^{-s}={\int }_0^{\infty }{x}^{s/2-1}{e}^{-\pi n^{2}x}dx$, obtained by substituting $x\mapsto x/(\pi n^2)$ in the Gamma integral. Summing over $n\ge 1$ for $\sigma >1$ and exchanging sum and integral (justified by absolute convergence) gives

The borrowed input is the theta transformation $\theta (1/x)=x^{1/2}\theta (x)$ for $\theta (x)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}x}$, itself a consequence of the Poisson summation formula applied to the Gaussian. Splitting the integral at $x=1$ and applying the transformation to the piece over $(0,1)$ yields

The integral converges for every $s$, because $\omega (x)$ decays like $e^{-\pi x}$; and the whole expression is visibly unchanged by $s\mapsto 1-s$. This both continues $\xi$ to $\mathbb{C}$ and proves the symmetry, with the two poles of $\xi$ at $s=0,1$ coming from the explicit fraction. The pole at $s=0$ is cancelled by the pole of $\Gamma (s/2)$ when one solves for $\zeta$.∎

Proof. For $\sigma >1$ non-vanishing follows from the theorem Euler product below: a convergent infinite product of nonzero factors is nonzero, which we verify there. For $\sigma <0$, use the unwound functional equation $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$. When $\sigma <0$ the argument $1-s$ has real part $>1$, so $\zeta (1-s)\ne 0$; also $\Gamma (1-s)$ has no zeros anywhere, and $2^s{\pi }^{s-1}$ never vanishes. So the zeros of $\zeta$ in $\sigma <0$ are exactly the zeros of $\sin (\pi s/2)$ there, namely $s=-2,-4,\dots$ (the zero of the sine at $s=0$ is cancelled by the pole of $\Gamma (1-s)$ at $s=1$ under the reflection, and $s=0$ is not a zero: $\zeta (0)=-1/2$).∎

Proof. Proof sketch — assumes the explicit formula and Hadamard product theory. Both directions run through the theorem Explicit formula below, which expresses $\psi (x)=x-{\sum }_{\rho }{x}^{\rho }/\rho -\log (2\pi )-\frac{1}{2}\log (1-x^{-2})$ as a sum over the nontrivial zeros. A zero $\rho =\beta +i\gamma$ contributes a term of modulus $x^{\beta }/|\rho |$. If every $\beta =\frac{1}{2}$ the sum is $O(\sqrt{x}{\log }^{2}x)$ after controlling the zero density, giving $\psi (x)=x+O(\sqrt{x}{\log }^{2}x)$, and partial summation converts this to the stated bound for $\pi (x)$. Conversely, a zero with $\beta >\frac{1}{2}$ forces $\psi (x)-x$ to exceed $x^{\beta -\epsilon }$ infinitely often — an oscillation theorem of Landau's type — contradicting the error bound. Neither direction is elementary: both need the Hadamard factorisation of $\xi$ to know that the zero sum converges at all.∎

Proof. Fix $s$ with $\sigma >1$ and let $P$ be a finite set of primes. Since $|p^{-s}|=p^{-\sigma }\le 2^{-\sigma }<1$, each factor is the sum of a convergent geometric series,

Multiply these together over $p\in P$. Each of the finitely many series converges absolutely, so the product may be expanded term by term and the resulting multiple series rearranged freely:

where $N(P)$ is the set of positive integers all of whose prime factors lie in $P$. Here both halves of unique factorization are used: existence of a factorization says every $n\in N(P)$ is hit, and uniqueness says it is hit exactly once, so no $n$ is counted twice and the right-hand sum is a genuine subsum of $\zeta (s)$, not an overcount.

Now take $P=P_y=\{p\le y\}$. Every integer $n\le y$ has all its prime factors $\le y$, so $N(P_y)\supseteq \{1,2,\dots ,\lfloor y\rfloor \}$, and therefore

The last sum is the tail of a convergent series, so it tends to $0$ as $y\to \infty$. Hence the partial products converge to $\zeta (s)$.

For absolute convergence of the product, note ${\sum }_p|p^{-s}|\le {\sum }_{n\ge 2}{n}^{-\sigma }<\infty$, which is the standard criterion for $\prod (1-a_p{)}^{-1}$ to converge absolutely to a nonzero limit. Non-vanishing follows: each factor is nonzero, and the limit of the partial products is nonzero because ${\sum }_p|p^{-s}|$ converges.∎

Proof. Suppose there were finitely many. Then the product ${\prod }_p(1-p^{-\sigma }{)}^{-1}$ is a finite product of real numbers, hence bounded as $\sigma \downarrow 1$ — its limit is the finite number ${\prod }_p(1-p^{-1}{)}^{-1}$. But by the theorem Euler product it equals $\zeta (\sigma )$, which by the theorem Convergence of the zeta series satisfies $\zeta (\sigma )\ge 1/(\sigma -1)\to \infty$. Contradiction.∎

Proof. Take logarithms in the theorem Euler product for real $\sigma >1$, which is legitimate since all factors are positive reals:

using $-\log (1-x)={\sum }_{k\ge 1}{x}^k/k$ for $|x|<1$. The remainder collects the terms $k\ge 2$ and is uniformly bounded:

So ${\sum }_p{p}^{-\sigma }=\log \zeta (\sigma )-R(\sigma )=\log \zeta (\sigma )+O(1)$. By the corollary Simple pole at $s=1$ with residue $1$, $\zeta (\sigma )\sim 1/(\sigma -1)$, so $\log \zeta (\sigma )=\log \frac{1}{\sigma -1}+o(1)$, which is the stated asymptotic.

Now let $\sigma \downarrow 1$. If ${\sum }_{p}1/p$ converged, say to $S$, then ${\sum }_p{p}^{-\sigma }\le S$ for every $\sigma >1$, since each term only decreases. But the right side tends to $+\infty$. Hence ${\sum }_{p}1/p$ diverges.∎

Proof. By the theorem Euler product, $1/\zeta (s)={\prod }_p(1-p^{-s})$. Expanding this product over $p\le y$ gives ${\sum }_{n\in N(P_y)}\mu (n)n^{-s}$: choosing the term $-p^{-s}$ from a subset $S$ of the primes and $1$ from the rest produces the squarefree integer $n={\prod }_{p\in S}p$ with sign $(-1{)}^{|S|}=\mu (n)$, and non-squarefree $n$ never arise, which is exactly the definition of $\mu$. The tail estimate of the theorem Euler product applies verbatim since $|\mu (n)|\le 1$.

For the second identity take the logarithmic derivative of $\zeta (s)={\prod }_p(1-p^{-s}{)}^{-1}$, permissible on $\sigma >1$ because the product converges locally uniformly and has no zeros:

and ${\sum }_p{\sum }_k\log p\cdot p^{-ks}$ is precisely ${\sum }_n\Lambda (n)n^{-s}$.∎

Proof. Count the exponent of $p$ in $N!$ by Legendre's formula: it is ${\sum }_{k\ge 1}\lfloor N/p^k\rfloor$. Hence

The $k=1$ terms give ${\sum }_{p\le N}\log p\lfloor N/p\rfloor =N{\sum }_{p\le N}\frac{\log p}{p}+O\left({\sum }_{p\le N}\log p\right)$, since $\lfloor N/p\rfloor =N/p+O(1)$. The terms $k\ge 2$ contribute at most

the last series converging by comparison with ${\sum }_m\log m/m^2$. By the theorem Chebyshev's bounds below, ${\sum }_{p\le N}\log p=O(N)$, so both error terms are $O(N)$ and

Finally Stirling — or just the integral comparison $\log (N!)={\sum }_{n\le N}\log n=N\log N-N+O(\log N)$ — gives $\log (N!)=N\log N+O(N)$. Dividing by $N$ yields ${\sum }_{p\le N}\frac{\log p}{p}=\log N+O(1)$, and passing from an integer $N$ to a real $x$ changes the sum by nothing and $\log x$ by $O(1)$.∎

Proof. Proof sketch — the analytic input is isolated in the section "How the theorem is proved". The chain is: (i) the proposition Chebyshev functions are interchangeable reduces the claim to $\psi (x)\sim x$; (ii) the proposition The Möbius function inverts zeta turns $\psi$ into a contour integral of $-{\zeta }^{\prime }/\zeta$; (iii) the theorem Non-vanishing on the line $\sigma =1$ shows the only pole of $-{\zeta }^{\prime }/\zeta$ on $\sigma \ge 1$ is the simple pole at $s=1$ of residue $1$; (iv) a Tauberian theorem converts that analytic fact into $\psi (x)\sim x$. Steps (i) and (iii) are proved in full below; (ii) and (iv) are the borrowed complex analysis.∎

Proof. $\psi$ versus $\theta$. By definition $\psi (x)-\theta (x)={\sum }_{k\ge 2}\theta (x^{1/k})$. The sum is empty once $x^{1/k}<2$, i.e. for $k>{\log }_{2}x$, and $\theta (y)\le y\log y$ trivially, so

Hence $\psi (x)\sim x⟺\theta (x)\sim x$.

$\theta$ versus $\pi$. One direction: $\theta (x)={\sum }_{p\le x}\log p\le \pi (x)\log x$. The other: for any $0<\epsilon <1$,

Dividing by $x$ and using $x^{1-\epsilon }\log x=o(x)$,

If $\theta (x)\sim x$, letting $x\to \infty$ and then $\epsilon \downarrow 0$ squeezes $\pi (x)\log x/x\to 1$. Conversely if $\pi (x)\sim x/\log x$ the same two inequalities force $\theta (x)/x\to 1$.∎

Proof. ($\Rightarrow$) Put $x=p_n$, so $\pi (p_n)=n$ and the hypothesis gives

Taking logarithms of $(\ast )$: $\log n+\log \log p_n-\log p_n\to 0$. Since $p_n\to \infty$, $\log \log p_n=o(\log p_n)$, so dividing by $\log p_n$ gives

Multiplying this by $(\ast )$ gives $n\log n/p_n\to 1$, i.e. $p_n\sim n\log n$.

($\Leftarrow$) Assume $p_n\sim n\log n$. Then $\log p_n=\log n+\log \log n+o(1)\sim \log n$, so

which is $(\ast )$. For general $x$, choose $n$ with $p_n\le x<p_{n+1}$, so $\pi (x)=n$ and

Both bounds tend to $1$, because $p_{n+1}/p_n\to 1$ and $\log p_{n+1}/\log p_n\to 1$ follow from $p_n\sim n\log n$. Hence $\pi (x)\log x/x\to 1$.∎

Proof. Using $\cos 2\vartheta =2{\cos }^2\vartheta -1$,

Proof. For real $\sigma >1$ and real $t$, take real parts in the logarithm of the Euler product. From the proof of the theorem Divergence of the sum of prime reciprocals,

Apply the lemma The three-four-one inequality with $\vartheta =kt\log p$ to each term of the combination

Exponentiating,

Now suppose $\zeta (1+it)=0$ for some $t\ne 0$. As $\sigma \downarrow 1$:

• $|\zeta (\sigma ){|}^3=O\left((\sigma -1{)}^{-3}\right)$, by the corollary Simple pole at $s=1$ with residue $1$;
• $|\zeta (\sigma +it){|}^4=O\left((\sigma -1{)}^4\right)$, since a zero at $1+it$ makes $\zeta (\sigma +it)=O(\sigma -1)$;
• $|\zeta (\sigma +2it)|=O(1)$, since $\zeta$ is holomorphic at $1+2it$ — here $t\ne 0$ is used, so that $1+2it\ne 1$.

The left side of $(†)$ is therefore $O(\sigma -1)$, which tends to $0$ and contradicts $(†)$.∎