limit of power of fraction of sums of sines Find the following limit:
$$\lim_{n\to\infty} \left(\frac{{\sin\frac{2}{2n}+\sin\frac{4}{2n}+\cdot \cdot \cdot+\sin\frac{2n}{2n}}}{{\sin\frac{1}{2n}+\sin\frac{3}{2n}+\cdot \cdot \cdot+\sin\frac{2n-1}{2n}}}\right)^{n}$$
I thought of some $\sin(x)$ approximation formula, but it doesn't seem to work.
 A: Let $f : [0, 1] \to [0, \infty)$ be of the class $C^1$ and not identically zero. Then by Mean Value Theorem, we have
$$ \sum_{k=1}^{n} f \left( \tfrac{2k}{2n} \right) = \sum_{k=1}^{n} \left( f \left( \tfrac{2k-1}{2n} \right) + f' (x_{n,k}) \frac{1}{2n} \right) $$
for some $x_{n,k} \in \left(\frac{2k-1}{2n}, \frac{2k}{2n} \right)$. Letting
$$ I_n = \frac{1}{n} \sum_{k=1}^{n} f \left( \tfrac{2k-1}{2n} \right) \quad \text{and} \quad J_n = \frac{1}{n} \sum_{k=1}^{n}f' (x_{n,k}),$$
We have
$$I_n \to I := \int_{0}^{1} f(x) \; dx \quad \text{and} \quad J_n \to J := \int_{0}^{1} f'(x) \; dx.$$
Therefore we obtain
$$ \left[ \frac{\sum_{k=1}^{n} f \left( \frac{2k}{2n} \right)}{\sum_{k=1}^{n} f \left( \frac{2k-1}{2n} \right)} \right]^{n} = \left( \frac{nI_n + \frac{1}{2}J_n}{n I_n} \right)^{n} = \left( 1 + \frac{1}{n}\frac{J_n}{2I_n} \right)^{n} \xrightarrow[n\to\infty]{} \exp \left( \frac{J}{2I} \right). $$
Now plugging $f(x) = \sin x$, the corresponding limit is $\exp \left( \frac{1}{2} \cot \frac{1}{2} \right)$.
A: $$\sum_{k=1}^n \sin \left(a + (k-1)d \right) = \dfrac{\sin(dn/2) \sin(a+d(n-1)/2)}{\sin(d/2)}$$
In your case, for the numerator $a = \dfrac{2}{2n}$ and $d = \dfrac{2}{2n}$. Hence, the numerator is $$ \dfrac{\sin(1/2) \sin(2/2n+2/2n \times (n-1)/2)}{\sin(1/2n)} = \dfrac{\sin(1/2) \sin((n+1)/(2n))}{\sin(1/2n)}$$
Similarly, the denominator gives $$\dfrac{\sin^2(1/2)}{\sin(1/2n)}$$
Hence, $$\dfrac{\dfrac{\sin(1/2) \sin((n+1)/(2n))}{\sin(1/2n)}}{\dfrac{\sin^2(1/2)}{\sin(1/2n)}} = \dfrac{\sin((n+1)/(2n))}{\sin(1/2)} = \dfrac{\sin(1/2) \cos(1/2n) + \cos(1/2) \sin(1/2n)}{\sin(1/2)}$$
The series expansion at $n= \infty$ gives us $$1 + \dfrac{\cot(1/2)}{2n} - \dfrac1{8n^2} + \mathcal{O}(1/n^3)$$
Hence, the desired limit is $$\lim_{n \rightarrow \infty} \left(\dfrac{\sin((n+1)/(2n))}{\sin(1/2)} \right)^n = \lim_{n \rightarrow \infty} \left(1 + \dfrac{\cot(1/2)}{2n} - \dfrac1{8n^2} + \mathcal{O}(1/n^3) \right)^n\\ = \exp \left(\dfrac12 \cot\left( \dfrac12 \right) \right)$$
A: Consider
$$z=\cos\frac{1}{2n}+i\sin\frac{1}{2n}$$
Then numerator and denominator are expressed respectively:
$$A_n=z^{2}+z^{4}+\cdots+z^{2n}=\frac{z^{2}\left(1-z^{2n+2}\right)}{1-z^{2}}$$
$$B_n=z+z^{3}+\cdots+z^{2n-1}=\frac{z\left(1-z^{2n+1}\right)}{1-z}$$
Separate real and imaginary parts
