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A zeta limit

Real Analysis
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Tolaso J Kos
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A zeta limit

#1

Post by Tolaso J Kos » Tue Apr 11, 2017 7:30 pm

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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Re: A zeta limit

#2

Post by Riemann » Sat May 26, 2018 1:28 pm

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 $\displaystyle \lim_{n \to +\infty} \frac{\zeta^{(n)}(0)}{n!} = -1$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: A zeta limit

#3

Post by Riemann » Sat May 26, 2018 1:31 pm

Using the above fact we get that $\zeta^{(n)}(0) \sim -n!$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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