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I came across the last part of this question and need some direction to find the solution. The original problem was to use de Moivre's theorem to show:

$$ \sin(2n+1)\theta = \binom{2n+1}{1}\cos^{2n}\theta \sin\theta - \binom{2n+1}{3}\cos^{2n-2}\theta \sin^3\theta +...+ (-1)^n\sin^{2n+1}\theta $$

Part (a),(b) and (c) of this problem are fine.

Part (c) of this question was:

Deduce that

$$ \cot^2 \bigg(\frac{\pi}{2n+1} \bigg) + \cot^2 \bigg(\frac{2\pi}{2n+1} \bigg) +...+ \cot^2 \bigg(\frac{n\pi}{2n+1} \bigg) = \frac{n(2n - 1)}{3} $$

I used Vieta's formula to prove part (c). However, I'm confused with part (d).

Part (d)

Use the fact that $\cot \theta < \dfrac{1}{\theta}$ for $0 < \theta < \dfrac{\pi}{2}$ to show that:

$$ \bigg( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} +...+ \frac{1}{n^2} \bigg) \frac{(2n+1)^2}{2n(2n-1)} > \frac{\pi^2}{6} $$

Any help appreciated.

Stephan
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    Hint: How can you use their fact/hint? – Calvin Lin Nov 24 '23 at 23:45
  • I also noticed the left hand side is increasing and the equation is true for n=1. You could use that if you wanted to prove it without trigonometry. – Jack Nov 25 '23 at 00:12
  • @Jack Are you sure? I believe the LHS is decreasing. The sequence is $\approx 4.5, 2.60, 2.22, 2.06, 1.97 , \ldots $. It takes a while to get down to $\pi^2/6 \approx 1.64$ (which I believe, but don't have proof as yet, is the limiting value) – Calvin Lin Nov 25 '23 at 00:29
  • You are right. I made a mistake when looking at it. – Jack Nov 25 '23 at 00:31

2 Answers2

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Rewrite the inequality (by cross multiplying) as

$$\sum_{k=1}^n [ \frac{ 2n+1 } { k \pi }]^2 > \frac{ n(2n-1) } { 3 } .$$

This makes it much easier to eyeball how to apply their fact/hint to part (c).

Calvin Lin
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  • Thanks Calvin. I assume since $\dfrac{1}{\theta^2} > \cot^2 \theta$ then it follows $$\sum_{k=1}^n \bigg[ \frac{ 2n+1 } { k \pi }\bigg]^2 > \frac{ n(2n-1) } { 3 } .$$ or $$\sum_{k=1}^n \bigg[ \frac{ 2n+1 } { k \pi }\bigg]^2 >\sum_{k=1}^n \cot^2\bigg(\frac{k\pi}{2n+1}\bigg) $$ – Stephan Nov 25 '23 at 03:54
  • @Stephan Essentially, but fill in that gap. – Calvin Lin Nov 25 '23 at 03:56
  • The 'gap' is to let $\theta = \dfrac{\pi}{2n+1}$? – Stephan Nov 25 '23 at 04:08
  • @Stephan I don't understand what you mean. Can you write it out in full? The gap is that we haven't proven the inequality (and merely restated it). – Calvin Lin Nov 25 '23 at 04:09
  • Here is my logic:

    Since $\frac{1}{\theta^2} > \cot^2\theta$, (this was given)then if $\theta = \frac{k\pi}{2n+1}$ where $k = 1, 2, 3,...$ $$ \frac{(2n+1)^2}{\pi^2} > \cot^2 \bigg(\frac{\pi}{2n+1}\bigg) $$

    $$ \frac{(2n+1)^2}{2^2\pi^2} > \cot^2 \bigg(\frac{2\pi}{2n+1}\bigg) $$

    $$ \frac{(2n+1)^2}{3^2\pi^2} > \cot^2 \bigg(\frac{3\pi}{2n+1}\bigg) $$

    e.t.c.

    Hence

    $$ \sum_{k=1}^{n} \bigg(\frac{2n+1}{k\pi}\bigg)^2 > \sum_{k=1}^{n} \cot^2 \Bigg( \frac{k\pi}{2n+1} \Bigg) $$

     – Stephan Nov 25 '23 at 04:41
  • @Stephan Yup, that's essentially it. – Calvin Lin Nov 25 '23 at 05:22
  • Thanks Calvin. You have been a great help! – Stephan Nov 25 '23 at 05:33
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Some background of this inequality: https://en.wikipedia.org/wiki/Basel_problem#Cauchy's_proof

Your question can be answered there and another similar inequality holds: \[ \frac{\pi^2}6\frac{2n(2n-1)}{(2n+1)^2}<\sum_{k=1}^n\frac1{k^2}<\frac{\pi^2}{6}\frac{4n(n+1)}{(2n+1)^2}. \]

ImbalanceDream
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