Estimation of a sum with cosecants

Real Analysis
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akotronis
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Estimation of a sum with cosecants

#1

Post by akotronis »

Show that $$\sum_{k=1}^{n-1}\csc\left(\frac{k\pi}{n}\right)=\frac{2}{\pi}n\ln n+\left(\frac{2\gamma}{\pi}-\frac{2\ln(\pi/2)}{\pi}\right)n+\mathcal O(1)\,.$$
Demetres
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Re: Estimation of a sum with cosecants

#2

Post by Demetres »

We begin by writing the sum in the form \[ 2\sum_{k=1}^{\lfloor (n-1)/2\rfloor} \csc\left(\frac{k\pi}{n}\right) + O(1) = 2\sum_{k=1}^{\lfloor (n-1)/2\rfloor} \frac{n}{k\pi} + 2\sum_{k=1}^{\lfloor (n-1)/2\rfloor} \left(\csc\left(\frac{k\pi}{n}\right) - \frac{n}{k\pi} \right) + O(1). \] The first term by the harmonic sum is equal to \[ \frac{2n}{\pi}\left( \log\left(\frac{n-1}{2} \right) + \gamma + O(1/n)\right) = \frac{2n \log{n}}{\pi} + \frac{(2\gamma - 2\log{2})n}{\pi}.\] For the second term writing \(f(x) = \csc(x) - \frac{1}{x}\) (with \(f(0):=0\)) we see that \[\frac{\pi}{n}\sum_{k=1}^{\lfloor (n-1)/2\rfloor} \left(\csc\left(\frac{k\pi}{n}\right) - \frac{n}{k\pi} \right)\] is the Riemann sum for \[ \int_0^{\pi/2} f(x) \, dx.\] Moreover, since \(f\) is increasing in \((0,\pi)\) the error in the approximation of the sum by the integral is at most \(\frac{\pi}{n}f(\pi/2) = O(1/n).\)

To compute the integral we observe that \[ F(x) = \log\left( \frac{\sin(x/2)}{x\cos{x/2}}\right) = \log(\sin(x/2)) - \log(\cos(x/2)) - \log(x)\] is an antiderivative of \(f\) and therefore \[ \int_0^{\pi/2} f(x) \, dx = F(\pi/2) - F(0) = -\log(\pi/2) + \log{2}.\] So the second term of the above expression is equal to \[ \frac{2n}{\pi}\left( -\log(\pi/2) + \log{2} \right) + O(1)\] and so the result follows.
akotronis
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Re: Estimation of a sum with cosecants

#3

Post by akotronis »

Thank you for the answer Demetres. The problem is from here.
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