# Error term when truncating series for $1/\zeta(s)$

Let $$s=\sigma+it$$, $$0\leq \sigma\leq 1$$, $$|t|\geq 1$$, say. Using Euler-Maclaurin, one can easily show that, for $$x\geq |t|$$, $$\zeta(s) = \sum_{n\leq x} \frac{1}{n^s} + \frac{x^{1-s}}{s-1} + O\left(\frac{1}{x^\sigma}\right).$$ (In fact, the implied constant is at most $$5/6$$.)

What about $$1/\zeta(s)$$? Can one also express $$\frac{1}{\zeta(s)} - \sum_{n\leq x} \frac{\mu(n)}{n^s}$$ as the sum of a leading term plus a much smaller error term, and, if so, what is the best known bound?

• Are you assuming RH? Obviously there are problems with zeros. Dec 13, 2019 at 17:35
• Say you are within a zero-free region. Dec 13, 2019 at 17:48
• If you're in a zero-free region you can use the usual bounds for partial sums of Mobius and then partial summation would do the rest. Dec 13, 2019 at 18:01
• I'm not sure that gives me quite white I am asking. Dec 13, 2019 at 18:11
• Section 14.25 of Titchmarsh, treats the case assuming RH. But do not get what you want.
– juan
Dec 13, 2019 at 20:00

I'd have a lousy suggestion, but it's too long for a comment so here goes. Maybe it's the wrong way to think with this problem, but my first idea would be to look for an approximate functional equation for the Möbius function, if such a concept makes sense. \ \ As far as I can see, Theorem 1 (page 47) of "The Approximate Functional Equation for a Class of Zeta-functions" Chandrasekharan and Narasimhan "almost" says, if I ignore logs,$$\begin {eqnarray*} &&\frac {1}{\zeta (s)}-\sum _{n\leq x}\frac {\mu (n)}{n^s}-\frac {\Delta (1-s)}{\Delta (s)}\sum _{n\leq y}\mu (n)n^{s-1} \\ &&\hspace {10mm}=\hspace {4mm}\frac {1}{2\pi i }\int _{\mathcal C}\frac {x^{z-s}dz}{\zeta (z)(s-z)} -x^{-s}\Big (A_{\lambda }^0(x)-S_0(x)\Big ) +\mathcal O\left (x^{-\sigma }+\frac {x^{1/2-\sigma }}{y^{1/2}}\right ) \end {eqnarray*}$$ where according to (46) of that paper essentially $$A_{\lambda }^0(x)-S_0(x)\ll 1$$ and where $$\mathcal C$$ is a closed contour containing all the singularities of the integrand so that the integral is something like $$\ll x^{1-\sigma }/t$$ and therefore the LHS above is $$\begin {eqnarray*} \ll \frac {x^{1-\sigma }}{t}+x^{-\sigma }+\frac {x^{1/2-\sigma }}{y^{1/2}} \end {eqnarray*}$$ which is an error quite a bit smaller than the main term for large $$t$$. I'm thinking of $$s$$ with around real part $$=1$$ here just to see what you can expect, but obviously in a zero-free region the integral bound above would be better.
\ \ But this theorem isn't applicable to $$1/\zeta (s)$$ because this function isn't holomorphic on a left half-plane :/ Perhaps it's possible to adapt the proof to accomodate for the infinitely many singularities of $$1/\zeta (s)$$.
• The theorem that you quote requires that $\Delta(s)\phi(s)$ has a holomorphic extension to the complement of a compact set, and the extension should tend to zero in every vertical strip (see page 32). Dec 31, 2019 at 0:17