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

1sin⁡(2π7)+1sin⁡(4π7)=1sin⁡(π7). \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

#Geometry #TelescopingSeries

Note by Calvin Lin
7 years ago

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Comments

Since

csc⁡2x=cot⁡x−cot⁡2x\csc 2x=\cot x-\cot 2x

the sum can be written as:

∑k=1n1sin⁡(2kπ2n+1−1)=∑k=1n(cot⁡(2k−1π2n+1−1)−cot⁡(2kπ2n+1−1))\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)

=cot⁡(π2n+1−1)−cot⁡(2nπ2n+1−1)=cot⁡(2n+1π2n+1−1)−cot⁡(2nπ2n+1−1)\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)

=−csc⁡(2n+1π2n+1−1)=csc⁡(π2n+1−1)\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=2n=2, it can be shown that:

csc⁡(2π7)+csc⁡(4π7)=csc⁡(π7)\displaystyle \csc\left(\frac{2\pi}{7}\right)+\csc\left(\frac{4\pi}{7}\right)=\csc\left(\frac{\pi}{7}\right)

Pranav Arora - 7 years ago

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Good job! I couldn't notice that!

Shaan Vaidya - 7 years ago

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

Anish Kelkar - 7 years ago

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

Calvin Lin Staff - 7 years ago
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