Growth rate of $\zeta^{(x)}(s)$

Lemma: For $k \in \mathbb{Z}^+$ , $(-1)^k \zeta^{(k)}(s) > 0$ .

Proof: $\displaystyle (-1)^k \zeta^{(k)}(s) = (-1)^k \sum_{n=1}^{\infty} \frac{(-\ln(n))^k}{n^s} = \sum_{n=1}^{\infty} \frac{\ln(n)^k}{n^s}$ . Since all the terms are positive, the sum is positive.

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The taylor series of $\zeta$ gives:

$\zeta(j) = \sum_{k=0}^\infty \frac{1}{k!} (j-s)^k \zeta^{(k)}(s)$

This gives

$\displaystyle \begin{aligned} \sum_{k=0}^\infty \frac{(1+\delta-s)^k}{k!} \zeta^{(k)}(s) &= \zeta(1+\delta), \quad \delta>0 \quad \text{-- (1)}\\ \sum_{k=0}^\infty \frac{(1-s)^k}{k!} \zeta^{(k)}(s) &= \lim_{\delta \rightarrow 0^+} \sum_{k=0}^\infty \frac{(1+\delta-s)^k}{k!} \zeta^{(k)}(s) \\ &= \lim_{\delta \rightarrow 0^+} \zeta(1+\delta) \quad \text{-- (2)} \end{aligned}$

Since $\zeta(s)$ diverges at $s=1$ but converges for all $s>1$ , $\text{(1)}$ converges while $\text{(2)}$ diverges. Let $a_k = \frac{(1+\delta-s)^k}{k!} \zeta^{(k)}(s)$ and $b_k = \frac{(1-s)^k}{k!} \zeta^{(k)}(s)$ . By the lemma, for $s>1+\delta$ , $a_k, b_k > 0$ for $k \in \mathbb{Z}^+$ as all the terms are positive. Hence, $\text{(1)}$ is absolutely convergent.

By the Ratio Test,

$\displaystyle \begin{aligned} \lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} \le &1 \le \lim_{k \rightarrow \infty} \frac{b_{k+1}}{b_k} \\ \lim_{k \rightarrow \infty} \frac{(1+\delta-s)\zeta^{(k+1)}(s)}{(k+1)\zeta^{(k)}(s)} \le &1 \le \lim_{k \rightarrow \infty} \frac{(1-s)\zeta^{(k+1)}(s)}{(k+1)\zeta^{(k)}(s)} \end{aligned}$

Limiting $\delta$ to $0^+$ , by Squeeze Theorem,

$\displaystyle \begin{aligned} &1 = \lim_{k \rightarrow \infty} \frac{(1-s)\zeta^{(k+1)}(s)}{(k+1)\zeta^{(k)}(s)} = \lim_{k \rightarrow \infty} \frac{(k+1)\zeta^{(k)}(s)}{(1-s)\zeta^{(k+1)}(s)} \\ &\lim_{k \rightarrow \infty} \frac{\zeta^{(k)}(s)}{\zeta^{(k+1)}(s)} = \lim_{k \rightarrow \infty} \frac{1-s}{k+1} = 0 \\ &\lim_{k \rightarrow \infty} \frac{\zeta^{(k)}(s)k}{\zeta^{(k+1)}(s)} = \lim_{k \rightarrow \infty} \frac{(k+1)\zeta^{(k)}(s)}{\zeta^{(k+1)}(s)} - \frac{\zeta^{(k)}(s)}{\zeta^{(k+1)}(s)} \\ &= 1-s-0 = 1-s \end{aligned}$

Hence the answer is when $s=2$ , $1-s = \boxed{-1}$ .

It is a well-known result that $\zeta^{(n)}(2) \sim (-1)^n n!$ (for $n \to \infty$ ); the proof is based on Julian's Lemma. It follows that the limit we seek is $\boxed{-1}$ .

The quoted result is a special case of $\zeta^{(n)}(s) \sim (-1)^n \frac{n!}{(s-1)^{n+1}}$ (for $\Re(s) >1$ )