A Riemann type sum

Real Analysis
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A Riemann type sum


Post by Riemann »

Justify the fact that:

$$\frac{1}{m} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \rightarrow \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x =\frac{1}{2}$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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