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Claim

$$ \lim_{n \to \infty} \left( \pi n \csc\left(\frac{\pi}{n}\right) - n^2 \right) = \sum_{k=1}^{\infty} \frac{1}{k^2} $$

Proof

Euler's infinite product $$ \sin(x) = x \prod_{k=1}^{\infty} \left(1 - \frac{x^2}{k^2 \pi^2}\right) $$

Set $x = \frac{\pi}{n}$ $$ \sin\left(\frac{\pi}{n}\right) = \frac{\pi}{n} \prod_{k=1}^{\infty} \left(1 - \frac{1}{k^2 n^2}\right) $$

Note

$$ \csc\left(\frac{\pi}{n}\right) = \frac{1}{\sin\left(\frac{\pi}{n}\right)} = \frac{n}{\pi} \prod_{k=1}^{\infty} \left(1 - \frac{1}{k^2 n^2}\right)^{-1} $$

Substituting $$ \begin{align*} \pi n \csc\left(\frac{\pi}{n}\right) - n^2 &= \pi n \left( \frac{n}{\pi} \prod_{k=1}^{\infty} \left(1 - \frac{1}{k^2 n^2}\right)^{-1} \right) - n^2 \\ &= n^2 \prod_{k=1}^{\infty} \left(1 - \frac{1}{k^2 n^2}\right)^{-1} - n^2 \\ &= n^2 \left( \prod_{k=1}^{\infty} \left(1 - \frac{1}{k^2 n^2}\right)^{-1} - 1 \right) \tag{*}\\ &\approx n^2 \left(1 + \sum_{k=1}^{\infty} \frac{1}{k^2 n^2} - 1 \right) \\ &\approx n^2 \sum_{k=1}^{\infty} \frac{1}{k^2 n^2}\\ &\approx \sum_{k=1}^{\infty} \frac{1}{k^2} \end{align*} $$

Taking the limits

$$ \lim_{n \to \infty} \left( \pi n \csc\left(\frac{\pi}{n}\right) - n^2 \right) = \lim_{n \to \infty} \sum_{k=1}^{\infty} \frac{1}{k^2} = \sum_{k=1}^{\infty} \frac{1}{k^2} \quad \blacksquare $$

Question

I applied the following approximations to the proof at $\left(*\right)$

$$ \overbrace{\prod_{k=1}^{\infty} \left(1 - \frac{1}{k^2 n^2}\right)^{-1} \approx \prod_{k=1}^{\infty} \left(1 + \frac{1}{k^2 n^2}\right)}^{(1 - a)^{-1} \approx 1 + a} \approx \overbrace{1 + \sum_{k=1}^{\infty} \frac{1}{k^2 n^2}}^{\prod_{k=1}^{\infty} (1 + a_k) \approx 1 + \sum_{k=1}^{\infty} a_k} $$

but I'm curious if those are the correct approximations to apply or do they invalidate the proof?

Pugs
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vengy
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    Vengy, in my opinion the approximation $(1-a)^{-1}\approx1+a$ invalidates the proof. – Angelo Nov 06 '24 at 19:05
  • Vengy, I think it would be easier to prove that both the limit and the series converge to $\frac{\pi^2}6$. Do you agree with me? If you want, I can prove it. – Angelo Nov 06 '24 at 19:21
  • Vengy, your proof looks simple, but in my opinion is not rigorous. If you are interested in proving that both the limit and the Basel series converge to the same value that is $\frac{\pi^2}6$, I can do it. – Angelo Nov 06 '24 at 19:29
  • Vengy, I would use the notable limits $\lim\limits_{x\to0}\dfrac{\sin x}x=1$ and $\lim\limits_{x\to0}\dfrac{x-\sin x}{x^3}=\dfrac16$. Are you interested in this approach ? – Angelo Nov 06 '24 at 23:19

3 Answers3

3

The expansion $$\csc z=\frac1z+\sum_{k=1}^\infty \frac{(-1)^k2z}{z^2-k^2\pi^2}$$ gives $$\pi n\csc(\tfrac\pi n)-n^2=\sum_{k=1}^\infty \frac{2(-1)^kn^2}{1-k^2n^2}$$ which is $2\eta(2)=\zeta(2)$ in the limit $n\to\infty$, by discrete version of dominated convergence theorem.

Bob Dobbs
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    Nice Bob! Another Basel Proof:

    $$ \overbrace{\lim_{n \to \infty} \left( \pi n \csc\left(\frac{\pi}{n}\right) - n^2 \right)}^{\frac{\pi^2}{6} \text{ Taylor Series}} \quad = \quad \overbrace{\lim_{n \to \infty} \sum_{k=1}^{\infty} \frac{2 (-1)^k n^2}{1 - k^2 n^2}}^{\sum_{k=1}^{\infty} \frac{1}{k^2} \text{ Basel Series}} \quad \implies \frac{\pi^2}{6} = \sum_{k=1}^{\infty} \frac{1}{k^2} $$

     – vengy Nov 06 '24 at 22:15
  • Bob, how would you prove that $\displaystyle\lim\limits_{n\to\infty}\sum\limits_{k=1}^\infty \frac{2(-1)^kn^2}{1-k^2n^2}=\sum\limits_{k=1}^\infty\lim\limits_{n\to\infty}\frac{2(-1)^kn^2}{1-k^2n^2};?$ – Angelo Nov 06 '24 at 23:27
  • @Angelo You can interchange the limit and the sum for partial sums. The tails are small. – Bob Dobbs Nov 06 '24 at 23:40
  • @BobDobbs, I know we can interchange the limit and the sum under certain hypotheses, for example, if the series is uniformly convergent. Please, could you write the steps in detail in order to prove that the limit and the sum can be interchanged ? – Angelo Nov 07 '24 at 12:44
  • $$ \small \lim_{n \to \infty} \sum_{k=1}^\infty \frac{2(-1)^k n^2}{1 - k^2 n^2} = \sum_{k=1}^\infty\frac{-2(-1)^k}{k^2} \cdot \underbrace{\frac{1}{1 - \frac{1}{k^2 n^2}}}{\approx , 1 , \text{as} , n \to \infty}=\sum{k=1}^\infty \frac{-2(-1)^k}{k^2} = 2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2} = 2 \eta(2)= \zeta(2) = \sum_{k=1}^{\infty} \frac{1}{k^2} $$ – vengy Nov 11 '24 at 15:14
3

You could make it shorter if you know that $\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi ^2}{6}$ $$\sin \left(\frac{\pi }{n}\right)=\frac{\pi }{n}-\frac{\pi ^3}{6n^3}+\frac{\pi ^5}{120n^5}+O\left(\frac{1}{n^7}\right)$$ Long division $$\csc \left(\frac{\pi }{n}\right)=\frac{n}{\pi }+\frac{\pi }{6 n}+\frac{7\pi ^3}{360 n^3}+O\left(\frac{1}{n^5}\right)$$ Then $$\pi n\,\csc \left(\frac{\pi }{n}\right)=n^2+\frac{\pi ^2}{6}+\frac{7 \pi ^4}{360 n^2}+O\left(\frac{1}{n^4}\right)$$ and $$\pi n\,\csc \left(\frac{\pi }{n}\right)-n^2=\frac{\pi ^2}{6}+\frac{7 \pi ^4}{360 n^2}+O\left(\frac{1}{n^4}\right)$$

Claude Leibovici
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Here is an elementary proof. Our starting point is the identity

$$ \begin{align*} \frac{n^2}{\sin^2(nx)} &= \sum_{k=0}^{n-1} \frac{1}{\sin^2\left(x + \frac{k \pi}{n}\right)} = \sum_{k=-\lfloor\frac{n-1}{2}\rfloor}^{\lceil\frac{n-1}{2}\rceil} \frac{1}{\sin^2\left(x + \frac{k \pi}{n}\right)} . \tag{1}\label{e:1} \end{align*} $$

This is easily proved by log-differentiating $ \prod_{k=0}^{n-1} \sin\left(x + \frac{k\pi}{n}\right) = 2^{1-n} \sin(nx) $ twice. Now plug $x = \frac{\theta}{n}$. Then by the dominated convergence theorem, as $n \to \infty$,

$$ \begin{align*} \frac{1}{\sin^2 \theta} = \sum_{k=-\lfloor\frac{n-1}{2}\rfloor}^{\lceil\frac{n-1}{2}\rceil} \frac{1}{n^2 \sin^2\left(\frac{\theta + k \pi}{n}\right)} \to \sum_{k=-\infty}^{\infty} \frac{1}{(\theta + k\pi)^2}. \end{align*} $$

Using this and invoking the dominated convergence theorem again,

$$ \begin{align*} \frac{\pi n}{\sin(\frac{\pi}{n})} - n^2 &= \pi n \left( \frac{1}{\sin(\frac{\pi}{n})} - \frac{n}{\pi} \right) \\ &= \frac{\pi n}{\frac{1}{\sin(\frac{\pi}{n})} + \frac{n}{\pi}} \left( \frac{1}{\sin^2(\frac{\pi}{n})} - \frac{n^2}{\pi^2} \right) \\ &= \frac{\pi n}{\frac{1}{\sin(\frac{\pi}{n})} + \frac{n}{\pi}} \sum_{k\in\mathbb{Z}\setminus\{0\}} \frac{1}{(\frac{\pi}{n} + k\pi)^2} \\ &\to \frac{\pi^2}{2} \sum_{k\in\mathbb{Z}\setminus\{0\}} \frac{1}{(k\pi)^2} \\ &= \sum_{k=1}^{\infty} \frac{1}{k^2}. \end{align*} $$

Sangchul Lee
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