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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.

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Thanks guys for your solutions. Sometimes it's really hard to choose the best solution because all are awesome. – user 1618033 May 29 '12 at 19:19
up vote 5 down vote accepted

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 $\exp \left( \frac{1}{2} \cot \frac{1}{2} \right)$.

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Great derivation! – Pedro Tamaroff May 29 '12 at 17:59
How did you think about this solution? – Pedro Tamaroff May 29 '12 at 19:06
@PeterTamaroff : It just reminded me of Riemann sum, so I attacked it in this way. – Sangchul Lee May 29 '12 at 19:33
I see. Maybe this will interest you. – Pedro Tamaroff May 29 '12 at 19:36

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

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$$\sum_{k=1}^n \sin \left(a + (k-1)d \right) = \dfrac{\sin(dn/2) \sin(a+d(n-1)/2)}{\sin(d/2)}$$

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 $$\lim_{n \rightarrow \infty} \left(\dfrac{\sin((n+1)/(2n))}{\sin(1/2)} \right)^n = \lim_{n \rightarrow \infty} \left(1 + \dfrac{\cot(1/2)}{2n} - \dfrac1{8n^2} + \mathcal{O}(1/n^3) \right)^n\\ = \exp \left(\dfrac12 \cot\left( \dfrac12 \right) \right)$$

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Do you have a proof for the sum? I guess it involves the sum formulas of the sine. – Pedro Tamaroff May 29 '12 at 17:19
You can use induction along with the sum formula for sine or rewrite the summation in terms of exponentials and then sum the geometric progression or use the sum formula for the $k$ term and the $(n-k)^{th}$ term. – user17762 May 29 '12 at 17:20

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