In a previous question, I derived the real and imaginary parts for $\zeta(s)$, where $s = \sigma + it$ and $\text{Re}(s) > 1$:
$$\text{Re}(\zeta(s)) = \sum_{n=1}^\infty \frac{\cos(t\log n)}{n^\sigma} \\ \text{Im}(\zeta(s)) = -\sum_{n=1}^\infty \frac{\sin(t\log n)}{n^\sigma} $$
These sums converge proportionally to the harmonic series; in other words, they converge very slowly.
How can we accelerate the convergence of these sums?
P.S.: I’m familiar with applying the Euler-Maclaurin formula to the Zeta Function, but I don’t see how this formula could apply to complex values. I tried plugging in $s = \sigma +it$ into that numerical algorithm, but things got complicated quickly.
I suspect your formulas are equivalent to
$$\Re(\zeta(s))=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N \left(n^{-(\Re(s)+i \Im(s))}+n^{-(\Re(s)-i \Im(s))}\right)\right)\tag{1}$$
and
$$\Im(\zeta(s))=\underset{N\to\infty}{\text{lim}}\left(-\frac{i}{2}\sum\limits_{n=1}^N \left(n^{-(\Re(s)+i \Im(s))}-n^{-(\Re(s)-i \Im(s))}\right)\right)\tag{2}$$.
in which case they're only valid for $\Re(s)>1$, and more terms will need to be evaluated the closer $s$ is to $1$ to maintain the same level of accuracy.
If you want accelerate the convergence near $s=1$, I suggest you consider using a globally convergent formula such as
$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum\limits_{n=0}^K \frac{1}{2^{n+1}}\sum\limits_{k=0}^n \frac{(-1)^k \binom{n}{k}}{(k+1)^s}\right)\tag{3}$$
(see Globally convergent series) which I believe can also be evaluated as
$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{\left(1-2^{1-s}\right) 2^{K+1}}\sum\limits_{n=0}^K \frac{(-1)^n}{(n+1)^s} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right)\tag{4}$$
(see formula (3) in my related question) which moves the exponentiation operation from the inner sum to the outer sum.
You can still use the relationships
$$\Re(\zeta(s))=\frac{1}{2} \left(\zeta(s)+\zeta\left(s^*\right)\right)\tag{5}$$
and
$$\Im(\zeta(s))=-\frac{i}{2} \left(\zeta(s)-\zeta\left(s^*\right)\right)\tag{6}$$
to separate out the real and imaginary parts if desired where $s^*$ represents the complex conjugate of $s$.
Another alternative would be to use a formula such as
$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{K^{1-s}}{s-1}+\sum\limits_{k=1}^K k^{-s}\right),\quad\Re(s)>0\tag{7}$$
which increases the rate of convergence near $s=1$ compared to the Dirichlet series
$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K k^{-s}\right),\quad\Re(s)>1\tag{8}$$
or better yet a formula such as
$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{(s+2 K-1)\, K^{-s}}{2\, (s-1)}+\sum\limits_{k=1}^{K-1} k^{-s}\right),\quad\Re(s)>-1\tag{9}$$
which converges more rapidly than formula (7) above for the entire critical strip.