Putnam 2019, Problem B-2

For all \(n \geq 1\), let
\[a_n = \sum_{k=1}^{n-1} \frac{\sin\left(\frac{(2k-1)\pi}{2n}\right)}{\cos^2\left(\frac{(k-1)\pi}{2n}\right) \cos^2\left(\frac{k\pi}{2n}\right)}.\]
Determine
\[\lim_{n \to \infty} \frac{a_n}{n^3}.\]

Solution:
The answer is $\frac{8}{\pi^3}$.

**Solution 1.** By the double angle and sum-product identities for cosine, we have
$$2\cos^2\left(\frac{(k-1)\pi}{2n}\right) - 2\cos^2\left(\frac{k\pi}{2n}\right) = \cos\left(\frac{(k-1)\pi}{n}\right) - \cos\left(\frac{k\pi}{n}\right)$$
$$= 2\sin\left(\frac{(2k-1)\pi}{2n}\right) \sin\left(\frac{\pi}{2n}\right),$$
and it follows that the summand in $a_n$ can be written as
$$\frac{1}{\sin\left(\frac{\pi}{2n}\right)} \left(-\frac{1}{\cos^2\left(\frac{(k-1)\pi}{2n}\right)} + \frac{1}{\cos^2\left(\frac{k\pi}{2n}\right)}\right).$$

Thus the sum telescopes and we find that
$$a_n = \frac{1}{\sin\left(\frac{\pi}{2n}\right)} \left(-1 + \frac{1}{\cos^2\left(\frac{(n-1)\pi}{2n}\right)}\right) = -\frac{1}{\sin\left(\frac{\pi}{2n}\right)} + \frac{1}{\sin^3\left(\frac{\pi}{2n}\right)}.$$

Finally, since $\lim_{x \to 0} \frac{\sin x}{x} = 1$, we have $\lim_{n \to \infty} (n \sin \frac{\pi}{2n}) = \frac{\pi}{2}$, and thus $\lim_{n \to \infty} \frac{a_n}{n^3} = \frac{8}{\pi^3}$.

**Solution 2.** We first substitute $n-k$ for $k$ to obtain
$$a_n = \sum_{k=1}^{n-1} \frac{\sin\left(\frac{(2k+1)\pi}{2n}\right)}{\sin^2\left(\frac{(k+1)\pi}{2n}\right) \sin^2\left(\frac{k\pi}{2n}\right)}.$$

We then use the estimate
$$\frac{\sin x}{x} = 1 + O(x^2) \quad (x \in [0,\pi])$$
to rewrite the summand as
$$\frac{\left(\frac{(2k+1)\pi}{2n}\right)}{\left(\frac{(k+1)\pi}{2n}\right)^2 \left(\frac{k\pi}{2n}\right)^2} \left(1 + O\left(\frac{k^2}{n^2}\right)\right)$$
which simplifies to
$$\frac{8(2k+1)n^3}{k^2(k+1)^2\pi^3} + O\left(\frac{n}{k}\right).$$

Consequently,
$$\frac{a_n}{n^3} = \sum_{k=1}^{n-1} \left(\frac{8(2k+1)}{k^2(k+1)^2\pi^3} + O\left(\frac{1}{kn^2}\right)\right)$$
$$= \frac{8}{\pi^3} \sum_{k=1}^{n-1} \frac{2k+1}{k^2(k+1)^2} + O\left(\frac{\log n}{n^2}\right).$$

Finally, note that
$$\sum_{k=1}^{n-1} \frac{2k+1}{k^2(k+1)^2} = \sum_{k=1}^{n-1} \left(\frac{1}{k^2} - \frac{1}{(k+1)^2}\right) = 1 - \frac{1}{n^2}$$
converges to 1, and so $\lim_{n \to \infty} \frac{a_n}{n^3} = \frac{8}{\pi^3}$.