# Calulate a limit involving $\zeta{(\zeta{(z)})}$

I'm currently trying to evaluate the following limit: $$\lambda=\lim_{z\to\infty}{\left[2^z-\left(\frac{4}{3}\right)^z-\zeta{(\zeta{(z)})}\right]}$$ A look at numerical approximations suggests, that $\lambda=1-\gamma$, but I have no idea how to tackle the iterated zeta function. I know that $\zeta(x)$ is asymptotical to $\frac{1}{1-x}$ when $x\to 1$, but this seems not to help here.

Any help will be highly appreciated!

• If you let $z$ approach $\infty$ in an arbitrary manner, the limit doesn't exist. Do you assume that $z$ is a positive real number, or maybe that $\operatorname{Re} z \to +\infty$? – Daniel Fischer Aug 18 '15 at 17:57

## 1 Answer

Assuming we're looking at the limit for $\operatorname{Re} z \to +\infty$, we have $\zeta(z) \to 1$, so the Laurent expansion

$$\zeta(s) = \frac{1}{s-1} + \gamma + O(s-1)\tag{1}$$

is relevant. We thus need to look at $\dfrac{1}{\zeta(z)-1}$ for large $\operatorname{Re} z$.

For $\operatorname{Re} z > 1$ and $k \geqslant 2$, define

$$R_k(z) = \sum_{n = k}^\infty \frac{1}{n^z}.$$

We will need bounds for $\lvert R_k(z)\rvert$: Letting $\sigma = \operatorname{Re} z$, we have

\begin{align} \lvert R_k(z)\rvert &\leqslant \sum_{n = k}^\infty \frac{1}{n^\sigma}\\ &< \frac{1}{k^\sigma} + \int_k^\infty \frac{dt}{t^\sigma}\\ &= \frac{1}{k^\sigma} + \frac{1}{(\sigma-1)k^{\sigma-1}}\\ &= \frac{1}{k^\sigma}\biggl(1 + \frac{k}{\sigma-1}\biggr). \end{align}

For $\lvert w\rvert \leqslant \frac{1}{2}$ we have

$$\biggl\lvert \frac{1}{1+w} - (1-w)\biggr\rvert = \frac{\lvert w\rvert^2}{\lvert 1+w\rvert} \leqslant 2\lvert w\rvert^2,$$

so with $\dfrac{1}{\zeta(z)-1} = \dfrac{2^z}{1 + 2^zR_3(z)}$ we obtain that

$$\biggl\lvert \frac{1}{\zeta(z)-1} - 2^z(1-2^zR_3(z))\biggr\rvert \leqslant 2\lvert 2^z\rvert\cdot\lvert 2^zR_3(z)\rvert^2 \leqslant 2\frac{2^{3\sigma}}{3^{2\sigma}}\biggl(1+\frac{3}{\sigma-1}\biggr)^2 \leqslant 8\cdot \biggl(\frac{8}{9}\biggr)^\sigma$$

for $\sigma = \operatorname{Re} z \geqslant 4$. Hence, using $R_3(z) = 3^{-z} + 4^{-z} + R_5(z)$,

\begin{align} \zeta(\zeta(z)) &= \frac{1}{\zeta(z)-1} + \gamma + O(\zeta(z)-1)\\ &= 2^z(1-2^zR_3(z)) + O\bigl((8/9)^{\operatorname{Re} z}\bigr) + \gamma + O\bigl(2^{-\operatorname{Re} z}\bigr)\\ &= 2^z - 2^{2z}\cdot 3^{-z} - 2^{2z}\cdot4^{-z} - 2^{2z}\cdot O\bigl(5^{-\operatorname{Re} z}\bigr) + \gamma + O\bigl((8/9)^{\operatorname{Re} z}\bigr)\\ &= 2^z - \biggl(\frac{4}{3}\biggr)^z - 1 + \gamma + O\bigl((8/9)^{\operatorname{Re} z}\bigr) \end{align}

for $\operatorname{Re} z \geqslant 4$, and therefore

$$\lim_{\operatorname{Re} z \to +\infty} \Biggl[ 2^z - \biggl(\frac{4}{3}\biggr)^z - \zeta(\zeta(z))\Biggr] = 1-\gamma.$$

If on the other hand we let $z_k = -2k$ for $k\in\mathbb{N}\setminus \{0\}$, then we see that $2^{z_k} \to 0$, $(4/3)^{z_k} \to 0$, and $\zeta(\zeta(z_k)) = \zeta(0)$, so

$$\lim_{k\to+\infty} \Biggl[2^{z_k} - \biggl(\frac{4}{3}\biggr)^{z_k} - \zeta(\zeta(z_k))\Biggr] = -\zeta(0) = \frac{1}{2} \neq 1-\gamma,$$

which shows that the limit as $z\to\infty$ of the expression doesn't exist.