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Taking the partial sum of the $\zeta $ function:

$$\zeta^H(s,k)=\sum_{n=0}^k \frac{1}{n^s}$$

and the partial product f the $\zeta $ function:

$$\zeta^P(s,j)=\prod_{i=0}^j \frac{1}{1-p_i^{-s}}$$

I have two questions:

(1) For $p$ being prime and $n,k,i\in\Bbb N$; do there exist closed functions for $\zeta^H(s,k)$ and $\zeta^P(s,j)$ to which the partial sum respectively the partial product converge?

(2) Does anybody know from a proof that would confirm that for a given $s$, if and if only $j,k\to\infty$ then: $$\zeta^H(s,k)-\zeta^P(s,j)\to0$$

Thanks for your help in advance.

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  • $\begingroup$ If the real part of $s$ is greater than 1, then both finite product and finite sum converge absolutely to $\zeta(s)$. $\endgroup$ – Sungjin Kim May 20 '13 at 20:04
  • $\begingroup$ sure, this is trivial. But my questions were about what happens before? $\endgroup$ – al-Hwarizmi May 21 '13 at 11:45
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For $s=1$ and positive nonzero integer $k$ you have the harmonic numbers. Their numerators are in the OEIS as sequence A001008 and their denominators as A002805.

That being said, there are other formulas that express your function $\zeta^H$. For instance, this paper by Borwein, Fee, Ferguson and Van Der Waall proposes the following (by denoting your function $\zeta^H (s,k)$ by $\zeta_N(s)=\displaystyle\sum_{n=1}^{N}n^{-s}$):

$$ \zeta(s) = \zeta_{N-1}(s) +N^{-s} + \frac{N^{1-s}}{a-1} + \sum_{n=1}^{\infty} \frac{B_{2n}}{(2n)!}\left( \prod_{j=0}^{2n-2} (s+j) \right) N^{1-s-2n} + R, $$

where the $B_n$ are the Bernouilli numbers.

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