Justify the fact that:
$$\frac{1}{m} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \rightarrow \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x =\frac{1}{2}$$
Welcome to mathimatikoi.org forum; Enjoy your visit here.
A Riemann type sum
A Riemann type sum
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Tags:
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 3 guests