Welcome to mathimatikoi.org forum; Enjoy your visit here.

A Riemann type sum

Real Analysis
Post Reply
User avatar
Riemann
Posts: 174
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

A Riemann type sum

#1

Post by Riemann »

Justify the fact that:

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

Tags:
Post Reply

Who is online

Users browsing this forum: No registered users and 1 guest