5
$\begingroup$

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.

  • 0
    Thanks guys for your solutions. Sometimes it's really hard to choose the best solution because all are awesome.2012-05-29

3 Answers 3

6

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

  • 0
    I see. Maybe [this](http://math.stackexchange.com/questions/150059/alternative-proof-of-the-limitof-the-quotient-of-two-sums) will interest you.2012-05-29
3

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

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

  • 0
    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.2012-05-29
3

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