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# A Trigonometric Geometric Sequence

Evaluate

$\sum_{k=1}^n \frac{1}{ \sin \left( \frac{ 2^k \pi} { 2^{n+1} -1} \right) }.$

In particular, show that

$\frac{ 1}{ \sin ( \frac{2\pi}{7} )} + \frac{1}{ \sin( \frac { 4 \pi } { 7} )} = \frac{1}{ \sin ( \frac{\pi}{7} ) }.$

This question is prompted by Daniel's Arc Co Tangent Sum.

Hint: Telescoping Series

Note by Calvin Lin
3 years, 4 months ago

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Since

$$\csc 2x=\cot x-\cot 2x$$

the sum can be written as:

$$\displaystyle \sum_{k=1}^n \frac{1}{\sin\left(\frac{2^k\pi}{2^{n+1}-1}\right)}=\sum_{k=1}^n \left(\cot\left(\frac{2^{k-1}\pi}{2^{n+1}-1}\right)-\cot\left(\frac{2^k\pi}{2^{n+1}-1}\right) \right)$$

$$\displaystyle =\cot\left(\frac{\pi}{2^{n+1}-1}\right)-\cot\left(\frac{2^n\pi}{2^{n+1}-1}\right)=\cot\left(\frac{2^{n+1}\pi}{2^{n+1}-1}\right)-\cot\left(\frac{2^n\pi}{2^{n+1}-1}\right)$$

$$\displaystyle =-\csc\left(\frac{2^{n+1}\pi}{2^{n+1}-1}\right)=\boxed{\csc\left(\dfrac{\pi}{2^{n+1}-1}\right)}$$

For $$n=2$$, it can be shown that:

$$\displaystyle \csc\left(\frac{2\pi}{7}\right)+\csc\left(\frac{4\pi}{7}\right)=\csc\left(\frac{\pi}{7}\right)$$

- 3 years, 4 months ago

Good job! I couldn't notice that!

- 3 years, 4 months ago

May be the part 2 of question is inspired from a question of IIT JEE 2011 . Am I correct ?

- 3 years, 4 months ago

I came across it in a different context, where it was given as a specific case of the first result.

Staff - 3 years, 4 months ago