Putnam_2019_B2

Putnam_2019_B2

Sure theres a nicer way but this is all I could come up with

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:loud_sound: :loud_sound: :loud_sound: EXCELLENT CONTENT :loud_sound: :loud_sound: :loud_sound:

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different approach.
we need to determine

\begin{align} \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\sin\left(\frac{(2k-1)\pi}{2n}\right)}{\cos^{2}\left(\frac{(k-1)\pi}{2n}\right)\cos^2\left(\frac{k\pi}{2n}\right)} &=\lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\sin\left(\frac{k\pi}{2n}+\frac{(k-1)\pi}{2n}\right)}{\cos^{2}\left(\frac{(k-1)\pi}{2n}\right)\cos^2\left(\frac{k\pi}{2n}\right)} \\&=\lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\sin\left(\frac{k\pi}{2n}\right)\cos\left(\frac{(k-1)\pi}{2n}\right)+\sin\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)}{\cos^{2}\left(\frac{(k-1)\pi}{2n}\right)\cos^2\left(\frac{k\pi}{2n}\right)} \\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\tan\left(\frac{k\pi}{2n}\right)+\tan\left(\frac{(k-1)\pi}{2n}\right)}{\cos\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)} \\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\left(\tan\left(\frac{k\pi}{2n}\right)+\tan\left(\frac{(k-1)\pi}{2n}\right)\right)\left(\tan\left(\frac{k\pi}{2n}\right)-\tan\left(\frac{(k-1)\pi}{2n}\right)\right)}{\cos\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)\left(\tan\left(\frac{k\pi}{2n}\right)-\tan\left(\frac{(k-1)\pi}{2n}\right)\right)} \\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\tan^2\left(\frac{k\pi}{2n}\right)-\tan^2\left(\frac{(k-1)\pi}{2n}\right)}{\sin\left(\frac{k\pi}{2n}\right)\cos\left(\frac{(k-1)\pi}{2n}\right)-\sin\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)} \\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\tan^2\left(\frac{k\pi}{2n}\right)-\tan^2\left(\frac{(k-1)\pi}{2n}\right)}{\sin\left(\frac{k\pi}{2n}-\frac{(k-1)\pi}{2n}\right)} \\ &= \lim_{n\to\infty} \frac{\csc\left(\frac{\pi}{2n}\right)}{n^3}\sum_{k=1}^{n-1}\tan^2\left(\frac{k\pi}{2n}\right)-\tan^2\left(\frac{(k-1)\pi}{2n}\right) \\ &= \lim_{n\to\infty} \frac{\csc\left(\frac{\pi}{2n}\right)}{n^3}\left(\sum_{k=2}^{n}\tan^2\left(\frac{(k-1)\pi}{2n}\right)-\sum_{k=1}^{n-1}\tan^2\left(\frac{(k-1)\pi}{2n}\right)\right) \\ &= \lim_{n\to\infty} \frac{\csc\left(\frac{\pi}{2n}\right)}{n^3}\left(\tan^2\left(\frac{(n-1)\pi}{2n}\right)-\tan^20\right) \\ &= \lim_{n\to\infty} \frac{\tan^2\left(\frac{(n-1)\pi}{2n}\right)}{n^3\sin\left(\frac{\pi}{2n}\right)} \ \ \ \ \ (*_1) \\ &= \lim_{n\to\infty} \frac{\tan^2\left(\frac \pi 2 -\frac{\pi}{2n}\right)}{n^3\left(\frac{\pi}{2n}\right)} \\ &= \frac 2 \pi \lim_{n\to\infty} \frac{\cot^2\left(\frac{\pi}{2n}\right)}{n^2} \\ &= \frac 2 \pi\lim_{n\to\infty} \frac{\cot\left(\frac{\pi}{2n}\right)\csc^2\left(\frac{\pi}{2n}\right)\left(\frac{\pi}{n^2}\right)}{2n} \ \ \ \ \ (*_2) \\ &= \lim_{n\to\infty} \frac{\cot\left(\frac{\pi}{2n}\right)\csc^2\left(\frac{\pi}{n}\right)}{n^3} \\ &= \lim_{n\to\infty} \frac{\cos\left(\frac{\pi}{2n}\right)}{\left(n\sin\left(\frac{\pi}{2n}\right)\right)^3} \\ &= \lim_{n\to\infty} \frac{1-\frac 1 2 \left(\frac{\pi}{2n}\right)^2}{\left(n\left(\frac{\pi}{2n}\right)\right)^3} \ \ \ \ \ (*_3),(*_4) \\ &= \frac{1}{\left(\frac{\pi}{2}\right)^3} \\ &=\frac{8}{\pi^3} \end{align}

noting that (*_1),(*_3),(*_4) are true due to small-angle approximations, and (*_2) is true due to l’hôpital’s rule

well that was quite \sqrt{1+\tan^2c} (get it)

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