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Find the following limit:

$$\lim_{n\to\infty}\frac{\sum_{k=1}^{n}\cos k+\sum_{k=1}^{n}\sin k}{\prod_{k=1}^{n}\cos k\sin k}$$

The numerator can be simplified by using Euler's formula and the sum of geometric series. I am struggling on the denomenator. How can we simplify that product? By the way, I don't even know whether or not this limit exist.

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    $\begingroup$ Use math.stackexchange.com/questions/17966/… and in the denominator, $\sin2k=2\sin k\cos k$ $\endgroup$ – lab bhattacharjee Jan 15 '14 at 7:21
  • $\begingroup$ Idle curiosity: Does it even have a convergent subsequence? Somehow, I doubt it. $\endgroup$ – Harald Hanche-Olsen Jan 15 '14 at 7:39
  • $\begingroup$ The denominator also simplifies by using Euler's formula. By the way, I also doubt that there is a limit. I better bet that, varying n, the term oscillates between minus infinity and plus infinity. $\endgroup$ – Claude Leibovici Jan 15 '14 at 8:00
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$\lim_{n\to\infty}\frac{\sum_{k=1}^{n}\cos k+\sum_{k=1}^{n}\sin k}{\prod_{k=1}^{n}\cos k\sin k}=\lim_{n\to\infty}\frac{\sum_{k=1}^{n}(cos k+sin k)}{\prod_{k=1}^{n}\cos k\sin k}=\lim_{n\to\infty}\frac{(cos 1+sin 2)+(cos 2+sin 2)+...+(cos n+sin n)}{(\cos 1\sin 1)(\cos 2\sin 2)...(\cos n\sin n)}\approx\lim_{n\to\infty}\frac{\cos n+\sin n}{\cos n\sin n}=\lim_{n\to\infty}(\frac{1}{\sin n}+\frac{1}{\cos n})$

I don't know how continue it

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