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?
 A: 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 $1/\zeta (s)$. 
