Quantcast
Channel: The cotangent sum $\sum_{k=0}^{n-1}(-1)^k\cot\Big(\frac{\pi}{4n}(2k+1)\Big)=n$ - MathOverflow
Browsing latest articles
Browse All 2 View Live

Answer by Iosif Pinelis for The cotangent sum...

The expression under the limit sign in question is just a Riemann sum for$$\int_0^{\pi/2} \frac12\,\Big(\frac1x-\cot x\Big)\,dx=\frac12\,\ln\frac\pi2,$$which therefore is the value of the limit.

View Article



The cotangent sum $\sum_{k=0}^{n-1}(-1)^k\cot\Big(\frac{\pi}{4n}(2k+1)\Big)=n$

On the Wolfram Research Reference page for the cotangent function (https://functions.wolfram.com/ElementaryFunctions/Cot/23/01/), I saw the following partial sum...

View Article
Browsing latest articles
Browse All 2 View Live




Latest Images