\subsection{Approximating $\zeta(s)$}
%Thus, the fact that $c_\vartheta$ is imaginary gives us no contradiction.
%We start with an application of Euler-Maclaurin.
\begin{lemma}\label{lem:abadeulmac}
Let $b>0$, $b\in \mathbb{Z} + \frac{1}{2}$. Then, for all $s\in \mathbb{C}\setminus \{1\}$ with $\Re s > 0$, 
\begin{equation}\label{eq:abak1}\sum_{n\leq b} \frac{1}{n^s} = \zeta(s) + \frac{b^{1-s}}{1-s} + s \int_b^\infty \left(\{y\}-\frac{1}{2}\right) \frac{dy}{y^{s+1}}.
\end{equation}
\end{lemma}
\begin{proof}
Assume first that $\Re s > 1$. By first-order Euler-Maclaurin and $b\in \mathbb{Z}+\frac{1}{2}$,
\[\sum_{n>b}\frac{1}{n^s} = \int_b^\infty \frac{dy}{y^s} + \int_b^\infty 
 \left(\{y\}-\frac{1}{2}\right) d\left(\frac{1}{y^s}\right).
\]
Here $\int_b^\infty \frac{dy}{y^s} = -\frac{b^{1-s}}{1-s}$ and $d\left(\frac{1}{y^s}\right) = 
- \frac{s}{y^{s+1}} dy$.
Hence, by $\sum_{n\leq b} \frac{1}{n^s} = \zeta(s) - \sum_{n>b} \frac{1}{n^s}$ for $\Re s > 1$,
$$\sum_{n\leq b} \frac{1}{n^s} = \zeta(s) + \frac{b^{1-s}}{1-s} +
\int_b^\infty 
 \left(\{y\}-\frac{1}{2}\right) \frac{s}{y^{s+1}} dy.$$
 Since the integral converges absolutely for $\Re s > 0$, both sides extend holomorphically to $\{s\in \mathbb{C}: \Re s>0, s\ne 1\}$; thus, the equation holds throughout that region.
\end{proof}

\begin{corollary}\label{cor:abadtoabsum}
Let $b>a>0$, $b\in \mathbb{Z} + \frac{1}{2}$. Then, for all $s\in \mathbb{C}\setminus \{1\}$ with
$\sigma = \Re s  > 0$,
$$\sum_{n\leq a} \frac{1}{n^s} = -\sum_{a<n\leq b} \frac{1}{n^s} + \zeta(s) + \frac{b^{1-s}}{1-s} + O^*\left(\frac{|s|}{2 \sigma b^\sigma}\right).$$
\end{corollary}
\begin{proof}
By Lemma \ref{lem:abadeulmac}, $\sum_{n\leq a} = \sum_{n\leq b} - \sum_{a<n\leq b}$,
$\left|\{y\}-\frac{1}{2}\right| \leq \frac{1}{2}$ and $\int_b^\infty \frac{dy}{|y^{s+1}|} = \frac{1}{\sigma b^\sigma}$.
\end{proof}

\begin{lemma}\label{lem:abadusepoisson}
Let $a,b\in \mathbb{R}\setminus \mathbb{Z}$, $b>a>0$.
Let  $s\in \mathbb{C}\setminus \{1\}$.
Define $f:\mathbb{R}\to\mathbb{C}$ by
 $f(y) = 1_{[a,b]}(y)/y^s$. Then 
$$\sum_{a<n\leq b} \frac{1}{n^s} =  \frac{b^{1-s} - a^{1-s}}{1-s} + \lim_{N\to \infty} \sum_{n=1}^N (\widehat{f}(n) + \widehat{f}(-n)) .$$
\end{lemma}
\begin{proof}
Since $a\notin \mathbb{Z}$, $\sum_{a<n\leq b} \frac{1}{n^s} = 
\sum_{n\in \mathbb{Z}} f(n)$.
By Poisson summation (as in \cite[Thm.~D.3]{MR2378655})
$$\sum_{n\in \mathbb{Z}} f(n) = \lim_{N\to \infty} \sum_{n=-N}^N \widehat{f}(n) = 
\widehat{f}(0) + \lim_{N\to \infty} \sum_{n=1}^N (\widehat{f}(n) + \widehat{f}(-n)) ,$$ 
where we use the facts that $f$ is in $L^1$, of bounded variation, and (by
$a,b\not\in \mathbb{Z}$)
continuous at every integer. Now
$$\widehat{f}(0) = \int_{\mathbb{R}} f(y) dy = \int_a^b \frac{dy}{y^s} = \frac{b^{1-s}-a^{1-s}}{1-s}.$$
\end{proof}

\begin{lemma}[Euler/Mittag-Leffler expansion for $\csc$ and $\csc^2$]\label{lem:abadeulmit}
Let $z\in \mathbb{C}$, $z\notin \mathbb{Z}$. Then
 \[ \frac{\pi}{\sin \pi z} =  \frac{1}{z} + 
 \sum_n (-1)^n\left(\frac{1}{z - n} + \frac{1}{z + n}\right),\quad
 \frac{\pi^2}{\sin^2 \pi z} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}.
 \]
\end{lemma}
We could prove these equations starting from Euler's product for $\sin \pi z$.
\begin{proof}
    Let us start from the Mittag-Leffler expansion $\pi \cot \pi s =
    \frac{1}{s} + \sum_n \left(\frac{1}{s-n} + \frac{1}{s+n}\right)$. 
    
    Applying the trigonometric identity $\cot u - \cot \left(u + \frac{\pi}{2}\right) = \cot u + \tan u = \frac{2}{\sin 2 u}$ with $u=\pi z/2$, and letting $s = z/2$, $s = (z+1)/2$, we see that
    \[\begin{aligned}\frac{\pi}{\sin \pi z} &= \frac{\pi}{2} \cot \frac{\pi z}{2} - \frac{\pi}{2} 
    \cot \frac{\pi (z+1)}{2} \\ &= \frac{1/2}{z/2} + 
    \sum_n \left(\frac{1/2}{\frac{z}{2} -n} + \frac{1/2}{\frac{z}{2} +n}\right)  -\frac{1/2}{(z+1)/2}
    -     \sum_n \left(\frac{1/2}{\frac{z+1}{2} -n} + \frac{1/2}{\frac{z+1}{2} +n}\right)\\
    &= \frac{1}{z} + \sum_n \left(\frac{1}{z - 2 n} + \frac{1}{z + 2 n}\right) -
    \sum_n \left(\frac{1}{z - (2 n - 1)} + \frac{1}{z + (2 n - 1)}\right)
    \end{aligned}\]
    after reindexing the second sum. Collecting the odd and even terms, we obtain our first equation.

    We may differentiate the expansion of $\pi \cot \pi z$ term-by-term 
    because it converges uniformly on compact subsets of $\mathbb{C}\setminus \mathbb{Z}$.
    By $\left(\pi \cot \pi z\right)' = - \frac{\pi^2}{\sin^2 \pi z}$ and
    $\left(\frac{1}{z\pm n}\right)' = -\frac{1}{(z\pm n)^2}$, we obtain our second equation.
\end{proof}

\begin{lemma}\label{lem:abadimpseri}
For $\vartheta\in \mathbb{R}$ with $0\leq |\vartheta|< 1$,
\[\sum_n\left(\frac{1}{(n-\vartheta)^3} + \frac{1}{(n+\vartheta)^3}\right)
\leq \frac{1}{(1-|\vartheta|)^3} + 2\zeta(3)-1.\]
\end{lemma}
\begin{proof} Since $\frac{1}{(n-\vartheta)^3} + \frac{1}{(n+\vartheta)^3}$ is even,
we may replace $\vartheta$ by $|\vartheta|$. Then we
rearrange the sum:
\[\sum_{n=1}^\infty \left(\frac{1}{(n-|\vartheta|)^3} + \frac{1}{(n+|\vartheta|)^3}\right) =\frac{1}{(1-|\vartheta|)^3} + \sum_{n=1}^\infty \left(\frac{1}{\left(n+1-|\vartheta|\right)^3} + \frac{1}{\left(n+|\vartheta|\right)^3}\right).\] 
We may write $(n+1-|\vartheta|)^3$, $(n+|\vartheta|)^3$
as $(n+\frac{1}{2}-t)^3$, $(n+\frac{1}{2} + t)^3$ for $t = |\vartheta|-1/2$.
Since $1/u^3$ is convex, $\frac{1}{(n+1/2-t)^3} + \frac{1}{(n+1/2+t)^3}$ reaches its maximum on $[-1/2,1/2]$ at the endpoints. Hence
\[\sum_{n=1}^\infty \left(\frac{1}{\left(n+1-|\vartheta|\right)^3} + \frac{1}{\left(n+|\vartheta|\right)^3}\right) \leq \sum_{n=1}^\infty \left(\frac{1}{n^3} + \frac{1}{(n+1)^3}\right) =  2 \zeta(3)-1.
\]
\end{proof}

\begin{lemma}\label{lem:abadsumas}
Let $s = \sigma + i \tau$, $\sigma\geq 0$, $\tau \in \mathbb{R}$, with
$s\ne 1$. 
Let $b>a>0$, $a, b\in \mathbb{Z} + \frac{1}{2}$, with  $a>\frac{|\tau|}{2\pi}$.
Define $f:\mathbb{R}\to\mathbb{C}$ by
 $f(y) = 1_{[a,b]}(y)/y^s$. Write $\vartheta = \frac{\tau}{2\pi a}$,
 $\vartheta_- = \frac{\tau}{2\pi b}$. Then
 $$\begin{aligned} \sum_n (\widehat{f}(n) + \widehat{f}(-n)) &= \frac{a^{-s} g(\vartheta)}{2 i} 
 - \frac{b^{-s} g(\vartheta_-)}{2 i} 
 + O^*\left(\frac{C_{\sigma,\vartheta}}{a^{\sigma+1}}\right)\end{aligned}$$
 with absolute convergence, 
where $g(t) =  \frac{1}{\sin \pi t} - \frac{1}{\pi t}$ for $t\ne 0$,
$g(0)=0$, and
\begin{equation}\label{eq:defcsigth}C_{\sigma,\vartheta}= \begin{cases} \frac{\sigma}{2} \left(\frac{1}{\sin^2\pi \vartheta} - \frac{1}{(\pi \vartheta)^2}\right) + \frac{|\vartheta|}{2\pi^2} \left(\frac{1}{(1-|\vartheta|)^3} + 2\zeta(3)-1\right) & \text{for $\vartheta\ne 0$,}\\ 
\sigma/6&  \text{for $\vartheta =  0$.}\end{cases}\end{equation}
\end{lemma}
\begin{proof}
By Proposition~\ref{prop:applem}, multiplying by $2$ (since $e(-n t)+e(n t) = 2 \cos 2\pi n t$), 
\begin{align}\widehat{f}(n) + \widehat{f}(-n) &= \notag
 \frac{a^{-s}}{2\pi i} \frac{(-1)^{n+1} 2\vartheta}{n^2 - \vartheta^2} -
 \frac{b^{-s}}{2\pi i} \frac{(-1)^{n+1} 2\vartheta_-}{n^2 - \vartheta_-^2}
 \\
 &+ \frac{a^{-\sigma-1}}{2\pi^2} O^*\left(\frac{\sigma}{(n-\vartheta)^2} + \frac{\sigma}{(n+\vartheta)^2} + \frac{|\vartheta|}{(n-\vartheta)^3} + \frac{|\vartheta|}{(n+\vartheta)^3}\right),\label{eq:abaderrcontrib}\end{align}
 where $\vartheta_- = \tau/(2\pi b)$. Note $|\vartheta_-|\leq |\vartheta|<1$.
 
%since $b^{-s} \frac{\tau}{2\pi b} = a b^{-s-1} \vartheta$ and 
% $n^2-\left(\frac{\tau}{2\pi b}\right)^2> n^2 - \vartheta^2$.
By the first equation in Lemma \ref{lem:abadeulmit}, 
\[\sum_n  \frac{(-1)^{n+1} 2 z}{n^2 - z^2}  = 
\frac{\pi}{\sin \pi z} - \frac{1}{z}
\]
for $z\ne 0$, while $\sum_n  \frac{(-1)^{n+1} 2 z}{n^2 - z^2} = \sum_n 0 = 0$ for
$z=0$.

Moreover, by Lemmas \ref{lem:abadeulmit} and \ref{lem:abadimpseri}, for $\vartheta\ne 0$,
\[\sum_n \left(\frac{\sigma}{(n-\vartheta)^2} + \frac{\sigma}{(n+\vartheta)^2}\right)\leq
\sigma\cdot \left(\frac{\pi^2}{\sin^2 \pi \vartheta} - \frac{1}{\vartheta^2}\right),\]
\[\sum_n \left(\frac{|\vartheta|}{(n-\vartheta)^3} + \frac{|\vartheta|}{(n+\vartheta)^3}\right)
\leq |\vartheta|\cdot \left( \frac{1}{(1-|\vartheta|)^3} + 2\zeta(3)-1\right).
\]
If $\vartheta=0$, then $\sum_n \left(\frac{\sigma}{(n-\vartheta)^2} + \frac{\sigma}{(n+\vartheta)^2}\right) = 2\sigma \sum_n \frac{1}{n^2} = \frac{\sigma \pi^2}{3}$, and then the quantity in \eqref{eq:abaderrcontrib} is bounded by 
 $\frac{a^{-\sigma-1}}{2\pi^2}\cdot \frac{\sigma \pi^2}{3} = \frac{\sigma/6}{a^{\sigma+1}}$.
\end{proof}
 \begin{proposition}\label{prop:dadaro}
 Let $s = \sigma + i \tau$, $\sigma\geq 0$, $\tau\in \mathbb{R}$, with $s\ne 1$.
 Let $a\in \mathbb{Z} + \frac{1}{2}$ with $a>\frac{|\tau|}{2\pi}$. Then
\begin{equation}\label{eq:abadlondie}\zeta(s) = \sum_{n\leq a} \frac{1}{n^s} - \frac{a^{1-s}}{1-s} + c_\vartheta a^{-s} + O^*\left(\frac{C_{\sigma,\vartheta}}{a^{\sigma+1}}\right),\end{equation}
where $\vartheta = \frac{\tau}{2\pi a}$, 
$c_\vartheta = \frac{i}{2}  \left(\frac{1}{\sin \pi \vartheta} - \frac{1}{\pi \vartheta}\right)$
for $\vartheta\ne 0$, $c_0 =0$, and $C_{\sigma,\vartheta}$ is as in \eqref{eq:defcsigth}.
\end{proposition}
If $0\leq \sigma<1$, then $\frac{a^{1-s}}{1-s} = \int_0^a \frac{dy}{y^s}$, and so 
\eqref{eq:abadlondie} can be read as expressing a difference $\sum_{n\leq a} - \int_0^a$ as a constant $\zeta(s)$ plus an error term.
\begin{proof} Assume first that $\sigma>0$. Let $b\in \mathbb{Z}+\frac{1}{2}$ with $b>a$, and define 
 $f(y) = \frac{1_{[a,b]}(y)}{y^s}$.  By Corollary~\ref{cor:abadtoabsum} and Lemma~\ref{lem:abadusepoisson},
$$\sum_{n\leq a} \frac{1}{n^s} = \zeta(s) + \frac{a^{1-s}}{1-s} - \lim_{N\to \infty} \sum_{n=1}^N (\widehat{f}(n) + \widehat{f}(-n)) + O^*\left(\frac{2 |s|}{\sigma b^\sigma}\right).$$
We apply Lemma~\ref{lem:abadsumas} to estimate 
$\lim_{N\to \infty} \sum_{n=1}^N (\widehat{f}(n) + \widehat{f}(-n))$. We obtain
\[\sum_{n\leq a} \frac{1}{n^s} = \zeta(s) + \frac{a^{1-s}}{1-s} - 
\frac{a^{-s} g(\vartheta)}{2 i} + O^*\left(\frac{C_{\sigma,\vartheta}}{a^{\sigma+1}}\right) + \frac{b^{-s} g(\vartheta_-)}{2 i}
+ O^*\left(\frac{2 |s|}{\sigma b^\sigma}\right),
\]
where $\vartheta_- = \frac{\tau}{2\pi b}$ and $g(t)$ is as in Lemma~\ref{lem:abadsumas}, and so $-\frac{g(\vartheta)}{2 i} = c_\vartheta$.
We let $b\to \infty$ through the half-integers, and obtain \eqref{eq:abadlondie},
since $b^{-\sigma}\to 0$, $\vartheta_-\to 0$ and  $g(\vartheta_-)\to g(0) = 0$  as $b\to \infty$.

Finally, the case $\sigma=0$ follows since all terms
in \eqref{eq:abadlondie} extend continuously to $\sigma=0$.
%To cover the case $\sigma=0$, use \eqref{eq:abadlondie}, and let $\sigma\to 0^+$. Since 
%$\frac{a^{1-s}}{1-s}$, $C_{\sigma,\vartheta}$ and $\frac{1}{a^{\sigma+1}}$ are continuous on 
%$\sigma$ for $a$, $\tau$ and $\vartheta = \tau/(2\pi a)$ fixed, we see \eqref{eq:abadlondie} also holds for $\sigma=0$.
\end{proof}

\begin{remark}
The term $c_\vartheta a^{-s}$ in \eqref{eq:abadlondie} does not seem to have been worked out before in the literature; the factor of $i$ in $c_\vartheta$ was a surprise. For the sake of comparison, let us note that, if $a\geq x$, then $|\vartheta|\leq 1/2\pi$, and so $|c_\vartheta|\leq |c_{\pm 1/2\pi}| = 0.04291\dotsc$ and
$|C_{\sigma,\vartheta}|\leq |C_{\sigma,\pm 1/2\pi}|\leq 0.176\sigma +0.246$. While $c_\vartheta$ is optimal, $C_{\sigma,\vartheta}$ need not be --
but then that is irrelevant for most applications.
\end{remark}