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 Post subject: A zeta limitPosted: Tue Apr 11, 2017 7:30 pm
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Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Let us denote with $\zeta$ the Riemann zeta function defined as $\zeta(0)=-\frac{1}{2}$. Let us also denote with $\zeta^{(n)}$ the $n$-th derivative of zeta. Evaluate the limit

$$\ell=\lim_{n \rightarrow +\infty} \frac{\zeta^{(n)}(0)}{n!}$$

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 Post subject: Re: A zeta limitPosted: Sat May 26, 2018 1:28 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 157
Location: Melbourne, Australia
We know that the function $\displaystyle f(z) \equiv \zeta(z) + \frac{1}{1-z}$ is a holomorphic function. The Taylor series around $0$ is

$$\zeta(z) + \frac{1}{1-z} = \sum_{n=0}^{\infty} \left( \frac{\zeta^{(n)}(0)}{n!} + 1 \right) z^n$$

which converges forall $z \in \mathbb{C}$ thus $\displaystyle \lim_{n \to +\infty} \frac{\zeta^{(n)}(0)}{n!} = -1$.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: A zeta limitPosted: Sat May 26, 2018 1:31 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 157
Location: Melbourne, Australia
Using the above fact we get that $\zeta^{(n)}(0) \sim -n!$.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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