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How to calculate the limit below? $$\lim_{n\to\infty}\{\frac{3}{n}\sum_{k=1}^n[1+8\sin^2(\frac{k\pi}{n})]^{-1}\}^{2^n}$$ Since I used Riemann integration to work out that $$\lim_{n\to\infty}[\frac{3}{n}\sum_{k=1}^n(1+8\sin^2(\frac{k\pi}{n}))^{-1}]=1$$, I've been trying to express $$\frac{3}{n}\sum_{k=1}^n(1+8\sin^2(\frac{k\pi}{n}))^{-1}=(1+\frac{C}{2^n}+o(\frac{1}{2^n})),\quad n\to+\infty$$ Can anyone render me some hints?

jimjim
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Drinzjeng Triang
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  • Nice question. If the exponent had been $n$ instead of $2^n$ the answer would have been simpler. It appears that the limit is $\infty$ but not sure. – Paramanand Singh Dec 29 '19 at 11:07
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    @Paramanand Singh .If the exponent had been n, the answer must be 1. – Drinzjeng Triang Dec 29 '19 at 11:09
  • Yeah. I think if the exponent were a suitable polynomial in $n$ then the limit would be finite and not equal to 1. Since $2^n$ tends to infinity much faster than polynomials the desired limit should be infinity. – Paramanand Singh Dec 29 '19 at 11:11
  • @Paramanand Singh, the key issue here misguiding your intuition on this problem is that, while in general Riemann sums tend to approximate integrals with error like $O(1/n^k)$, they are immensely better on periodic smooth functions like we have here. So this problem may well have some answer other than $\infty$ or $0.$ – Ragib Zaman Dec 29 '19 at 12:05
  • See "The Exponentially Convergent Trapezoidal Rule" by Trefethen and Weideman. – Ragib Zaman Dec 29 '19 at 12:06
  • @RagibZaman .Thanks for your recommendation, but haven't learned complex-analysis yet. I can only understand the solution with real-analysis. – Drinzjeng Triang Dec 29 '19 at 12:25
  • The link for a PDF about "The Exponentially Convergent Trapezoidal Rule" :https://pdfs.semanticscholar.org/cb7f/d8d180e9281f32d8712f7ee075e0561779fd.pdf – Drinzjeng Triang Dec 29 '19 at 12:26
  • @RagibZaman: well that's exactly the problem I was facing. The function here is analytic and periodic with period $1$ so all the differences $f^{(k)} (1)-f^{(k)}(0)$ were vanishing. I will read the paper in detail as it appears interesting. – Paramanand Singh Dec 29 '19 at 13:35

2 Answers2

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Now to prove Claude's observation and stated explicitly by the proposer in a comment. Start with the known (Hansen, Table of Series and Products, eq. 91.1.17)

$$ \prod_{k=0}^{n-1} \sinh^2{y} + \sin^2(x+k\,\pi/n) = 2^{1-2n}\big(\cosh{(2ny)} - \cos{(2nx)} \big)$$ Set x=0, note that sum can start with 1 and end with n, and take the logarithmic derivative, $$ 2\, \cosh{y}\,\sinh{y} \, \sum_{k=1}^n \frac{1}{\sinh^2{y} + \sin^2(k\,\pi/n)} = 2n \frac{\sinh{2ny}}{\cosh{(2ny)} - 1} $$

Solving $1/\sinh^2{y} = 8 \implies y=\log{2}/2.$ Algebra and hyperbolic trig ID completes the proof of

$$ \sum_{k=1}^n \frac{1}{1 + 8\sin^2(k\,\pi/n)}=\frac{n}{3}\,\coth{\big(\frac{n}{2}\log{2}\big)}=\frac{n}{3} \frac{2^n+1}{2^n-1} $$

user321120
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From what it seems$$\frac{3 }{n}\sum _{k=1}^n \frac{1}{1+8 \sin ^2\left(\frac{\pi k}{n}\right)}=\frac{2^n+1}{2^n-1}=1+\frac 2 {2^n}+ \cdots$$

So, now we consider $$S=\left(\frac{2^n+1}{2^n-1}\right)^{2^n}$$ Let $x=2^n$ making $$S=\left(\frac{x+1}{x-1}\right)^{x}\implies \log(S)=x \log\left(\frac{x+1}{x-1}\right)=x \log\left(1+\frac{2}{x-1}\right)$$ So, by Taylor $$\log(S)=2+\frac{2}{3 x^2}+\frac{2}{5 x^4}+O\left(\frac{1}{x^6}\right)$$ $$S=e^{\log(S)}=e^2+\frac{2 e^2}{3 x^2}+\frac{28 e^2}{45 x^4}+O\left(\frac{1}{x^6}\right)\qquad \text{with}\qquad x=2^n$$

Trying for $n=4$, the exact result is $$\frac{48661191875666868481}{6568408355712890625}\approx 7.40836885$$ while the above truncated formula gives $$\frac{739207 }{737280}e^2 \approx 7.40836859$$

Claude Leibovici
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