# How to make the real and imaginary parts of the Riemann Zeta Function converge faster?

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.

• The first idea I’m thinking about is to consider $(1-2^{1-s})\zeta(s)=\sum_{n \geq 1}{\frac{1}{(2n-1)^s}-\frac{1}{(2n)^s}}$. This converges in $O(n^{s+1})$ so slightly faster (although, of course, it’s still rather slow). Jun 9 at 16:27

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.

• Excuse me, because I’m still pretty new to complex analysis. I’m confused about why you still have the $s$ left over in both the real and imaginary parts, which would have an $it$ term in it. Wouldn’t that mean that both the real and imaginary parts would be complex, assuming $t$ is not zero? Jun 9 at 17:12
• @Mailbox I'm not sure which formulas you're looking at (formulas (1) and (2) or formulas (5) and (6)?), but assuming $s=\sigma+i\,t$ where $\sigma,t\in\mathbb{R}$, then $\Re(s)=\sigma$ and $\Im(s)=t$. Formulas (1) and (2) (and I suspect your formulas as well) are based on formulas (5) and (6). Jun 9 at 17:16
• @Mailbox Formulas (5) and (6) are based on $\Im(\zeta (s))=-\Im\left(\zeta \left(s^*\right)\right)$ where $s^*$ represents the complex conjugate of $s$, so if $s=\sigma+i\,t$ then $s^*=\sigma-i\,t$. Jun 9 at 19:50
• Let’s take formula (5), for example. Substituting $s = \sigma + it$, we get $\Re(\zeta(s)) = \frac12 \zeta(\sigma + it) + \frac12 \zeta(\sigma - it)$. The outputs of $\zeta(z)$ will always be a complex number, so we’ll let $\frac12 \zeta(\sigma + it) = a + bi$ and $\frac12 \zeta(\sigma - it) = c + di$. This means that $\Re(\zeta(s)) = (a + c) + i(b + d)$. But this is nonsensical to me! How can the real part of the zeta function be a complex number—how could it have an imaginary part? Jun 9 at 21:28
• @Mailbox Sorry, I mentioned $\Im(\zeta (s))=-\Im\left(\zeta \left(s^*\right)\right)$ but forgot to mention $\Re(\zeta(s))=\Re\left(\zeta\left(s^*\right)\right)$. Jun 9 at 22:49