Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 / A (Analytic Number Theory)

absolute convergence depends only on $\sigma =\mathrm{Re}s$, since $|a_n{n}^{-s}|=|a_n|n^{-\sigma }$. Let $A$ be the set of real $\sigma$ for which $\sum |a_n|n^{-\sigma }$ converges. Comparison with a convergent member shows that $A$ is upward closed. The hypotheses give a member of $A$ and a real number outside $A$, so its infimum ${\sigma }_0$ is finite. If $\sigma >{\sigma }_0$, some $\tau \in A$ lies below $\sigma$, giving convergence by comparison; if $\sigma <{\sigma }_0$, convergence would contradict the definition of the infimum. Thus ${\sigma }_0=\inf A$ is the abscissa of absolute convergence of a Dirichlet series. This argument leaves convergence on the boundary open.

Partition $1,\dots ,n$ according to $d=n/\gcd (k,n)$. Writing $k=(n/d)j$ identifies the part indexed by $d\mid n$ with the integers $1\le j\le d$ coprime to $d$. There are $\varphi (d)$ of them. Counting the entire partition proves the totient divisor-sum identity

We use the multiplication theorem for absolutely convergent Dirichlet series: their product is the Dirichlet series of the Dirichlet convolution of their coefficients. absolute convergence justifies rearrangement of the double sum. Here $0<\varphi (n)\le n$, so its series converges absolutely for $\sigma >2$, andThe Euler product implies $\zeta (s)\ne 0$ for $\sigma >1$. Division gives

We finally use the simple pole of the Riemann zeta function at one, of residue one, and $0<\zeta (2)<\infty$. The quotient therefore tends to $+\infty$ as real $s\downarrow 2$. If $\sum \varphi (n)n^{-2}$ converged, its nonnegative terms would dominate the series at every real $s>2$, giving a uniform finite bound and a contradiction. Thus the series diverges at two; comparison then excludes absolute convergence at every $\sigma <2$. Combined with the convergence already proved, the abscissa is exactly two. This is the totient Dirichlet series and its absolute convergence boundary.

summation by parts, with $S(x)={\sum }_{n\le x}\chi (n)$, gives for an integer $N$The boundary term tends to zero when $\mathrm{Re}s>0$ because $S$ is bounded. In the original region of absolute convergence this provesFor a compact subset of $\mathrm{Re}s>0$, choose $\delta >0$ below all its real parts and bound $|s|$ there. The integral tail is bounded by a constant times ${\int }_N^{\infty }{x}^{-1-\delta }dx$, uniformly on that compact set. Its truncated integrals are holomorphic function in $s$, so locally uniform convergence proves a analytic continuation throughout $\mathrm{Re}s>0$. This is holomorphy of a Dirichlet series from bounded partial sums.

The Euler product, valid for $\mathrm{Re}s>1$, isExpand the logarithms of these absolutely convergent Euler products. Writing $\theta =mt\log p$, the logarithm of the product in the desired inequality isFor odd $p$, $\chi (p{)}^m$ is $1$ or $-1$. In the first case the numerator is the given nonnegative trigonometric polynomial. In the second replace $\theta$ by $\theta +\pi$, which changes the sign of $\cos \theta$ and leaves $\cos 2\theta$ unchanged. At $p=2$ the numerator is $3+\cos 2\theta \ge 2$. Every summand is nonnegative. Exponentiation provesThis is the three-four-one inequality for Euler products applied to the real character modulo four.

Suppose $t\ne 0$ and $L(1+it)=0$. Holomorphy gives $L(\sigma +it)=O(\sigma -1)$ as $\sigma \downarrow 1$. The only pole of the Riemann zeta function is at one, so $\zeta (\sigma +2it)$ remains bounded, while $\zeta (\sigma {)}^3=O((\sigma -1{)}^{-3})$. The displayed product is then $O(\sigma -1)$, contradicting its lower bound. Hence $L(1+it)\ne 0$ for $t\ne 0$.

At $t=0$ the last zeta factor also has a pole, so that contradiction no longer follows from the inequality. Instead evaluate the convergent alternating series:For rigor, integrate the finite geometric identity for $(1+x^2{)}^{-1}$; the remainder has absolute integral at most $1/(2N+3)$ and tends to zero. summation by parts above identifies this value with the analytic continuation at one.

Write $s=\sigma +it$; the source's rectangle notation $|\mathrm{Im}(t)|\le T$ is interpreted as the intended height restriction $|t|\le T$. We use the standard facts that $\zeta$ has no zeros on $\sigma \ge 1$, has a simple pole at one, and is meromorphic with no other pole. The function $(s-1)\zeta (s)$ is holomorphic function and nonzero on the compact set $\{1+it:|t|\le T\}$, including its removable value one at $s=1$. It remains nonzero in a neighborhood of this compact set. Together with nonvanishing to its right this gives some $0<{\alpha }_T<1$ such that the whole closed rectangle ${\alpha }_T\le \sigma \le 2$, $|t|\le T$ contains no zeta zero. The pole at one is harmless: $1/\zeta$ extends holomorphically there with value zero.

The linearly smoothed Perron kernel, for $y>0$ and $c>0$, isFor $y>1$ shift the contour left across the simple poles at zero and minus one, whose residues are $y$ and $-1$. For $y<1$ shift right, crossing no pole. At $y=1$ the integral is zero, also obtained by direct integration of $1/s-1/(s+1)$. The quadratic denominator gives absolute integrability on the original vertical line.

For $c>1$ use the absolutely convergent Dirichlet series $1/\zeta (s)=\sum \mu (n)n^{-s}$. Interchanging the sum and the contour integral is justified by $\sum |\mu (n)|n^{-c}<\infty$ and the integrable denominator. The term for $n$ is $n$ times the kernel at $y=x/n$, hence $(x-n)$ when $n<x$ and zero otherwise. Consequently

Here is a contour argument that needs only the stated bound, with no unproved uniform estimate left of one. Take $x>e$, $c=1+1/\log x$, and first hold $T\ge 2$ fixed. On the two tails $|t|>T$, the given bound and $x^{c+1}=e{x}^2$ giveShift the remaining segment to $\sigma ={\alpha }_T$. The integrand is holomorphic function inside the rectangle: the zero of $1/\zeta$ at one cancels the apparent zeta pole, and $0,-1$ lie outside. On the fixed left side its integral is $O_T(x^{1+{\alpha }_T})=o_T(x^2)$. On the two horizontal sides, $1/(\zeta (s)s(s+1))$ is bounded by some finite constant depending on $T$. Therefore their integrals are bounded byIt follows thatNow let $T\to \infty$; the integrable tail tends to zero. ThusThe order of limits matters: first $x\to \infty$ for each fixed rectangle, then $T\to \infty$. This is linearly smoothed Möbius cancellation from finite zero-free rectangles.