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A question from some Russian book, about different summations an integrals, by Prudnikhov, Brichkhov and Marichev.

Page 746, 20, they write:

$$ \sum_{k=1}^{\infty}\prod_{m=1}^{2k}\text{ctg}\frac{m\pi}{2k+1}=\frac{\pi}{4}-1 $$

Without proof, actually the whole book is with no single proof.

How could I start proving that, what is product of $\prod_{m=1}^{2k}\text{ctg}\frac{m\pi}{2k+1}$?

Thanks!

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1 Answer 1

up vote 5 down vote accepted

I believe we can show that

$$\prod_{m=1}^{2k} \cot \frac{m\pi}{2k+1} = (-1)^k \frac{1}{2k+1}$$

Using Chebyshev Polynomials $\displaystyle T_n(x)$ (of degree $\displaystyle n$, and leading coefficient $\displaystyle 2^{n-1}$), which satisfy

$$T_{n}(\cos x) = \cos (n x)$$

and

$$T_{2n+1}(\sin x) = (-1)^k\sin ((2n+1) x)$$

We get the product of roots of $\displaystyle T_{2k+1}(x) = -1$, to find $\displaystyle \prod_{m=1}^{2k+1} (-1)^k\cos \frac{m\pi}{2k+1}$

(Note that its roots are $\displaystyle \cos (\frac{(2r+1)\pi}{2k+1})$ and $\displaystyle -\cos (\frac{2r\pi}{2k+1})$)

which gives us $$\prod_{m=1}^{2k} \cos \frac{m\pi}{2k+1} = \frac{(-1)^k}{2^{2k}}$$

To find $\displaystyle \prod_{m=1}^{2k} \sin \frac{m\pi}{2k+1}$, we need to find the product of roots of $\displaystyle \frac{T_{2k+1}(x)}{x} = 0$.

We can prove that the coefficient of $\displaystyle x$ in $\displaystyle T_{2k+1}(x)$ is $\displaystyle 2k+1$ and that would give us

$$\prod_{m=1}^{2k} \sin\frac{m\pi}{2k+1} = \frac{2k+1}{2^{2k}}$$

Thus

$$\prod_{m=1}^{2k} \cot\frac{m\pi}{2k+1} = \frac{(-1)^k}{2k+1}$$

Since $$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} = \int_{0}^{1} \frac{1}{1+x^2} \text{ dx} = \frac{\pi}{4}$$

we are done.

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Thank you very much! I'll study your proof... –  Salech Alhasov Feb 26 '12 at 21:58

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