Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 3

For $\sigma >1$, summation over the intervals $[n,n+1)$ givesSince $\lfloor u\rfloor =u-\{u\}$,The integral converges absolutely and defines a holomorphic function for $\sigma >0$. This formula therefore gives the Meromorphic continuation of the Riemann zeta function to the right half-plane, with only a simple pole at $s=1$ and residue one.

For $\sigma >1$, both relevant Dirichlet series converge absolutely, so their product may be rearranged:The inner sum is one when $n=1$ and zero otherwise by Möbius inversion. HenceThis is also the reciprocal of the absolutely convergent Euler product.

For $\sigma >1$, the three-four-one zero-free-region argument starts fromApplying this termwise togives

Suppose $\rho =\beta +i\gamma$ is a zero with $|\gamma |\ge 4$. In the supplied Local partial-fraction expansion of the Riemann zeta logarithmic derivative, every nearby zero contributes a nonnegative real part at $\sigma +i\gamma$ when $\sigma >1$. Keeping the term from $\rho$ givesAt height zero the pole at one givesand the same local expansion at height $2\gamma$ givesSubstitution yieldsSet $\sigma =1+\eta /\log |\gamma |$, first choosing a sufficiently small absolute $\eta >0$. If $\beta >1-c/\log |\gamma |$, the left side is at least $4\log |\gamma |/(\eta +c)$. Choosing $c>0$ sufficiently small contradicts the last inequality. Therefore

The truncated Perron formula says that, for $c>1$ and $T\ge 2$,where an endpoint has half weight and the error is controlled by

Apply this with $a_n=\mu (n)$, $c=1+1/\log x$, andPart (b) makes the integrand $x^s/(s\zeta (s))$. Move the contour tousing a contour that stays inside the Zero-free region of the Riemann zeta function near small $|t|$. The estimates supplied in the question give $1/\zeta (s)\ll \log (|t|+3)$ on the new contour. Its vertical segment is thereforeafter decreasing $c_3$. The horizontal segments and the Perron truncation error areNo residue is crossed because $1/\zeta (s)$ has a zero, rather than a pole, at $s=1$. Thus the Mertens function satisfiesfor some absolute $c>0$.