# Prove that sum_(k=1)^n 1/(sin2^kx)=cot x - cot 2^nx for every x ne (kpi)/2^k, x in RR, n in NN^+?

Dec 1, 2016

Not for this sum but for a similar one. Please see explanation.

#### Explanation:

I get a similar result for

$\sum \frac{1}{\sin} ^ \left({2}^{k}\right) x$

=(1-1/sin^(2^(n+1))x))/(1-1/sin^2x)-1,

using $1 + X + {X}^{2} + \ldots + {X}^{n} = \frac{1 - {X}^{n + 1}}{1 - X}$

$= - {\tan}^{2} x \left(1 - {\csc}^{{2}^{n + 1}} x\right) - 1$

$= {\sec}^{2} x \left({\csc}^{{2}^{n}} x - 1\right)$.

I admit that I don't get any idea, for solving the given problem.

Dec 1, 2016

See below.

#### Explanation:

Always is worth to read Ramanujan's Third Notebook.

With the identity

$\cot \left(x\right) = \cot \left(\frac{x}{2}\right) - \csc \left(x\right)$ we get

$\frac{1}{\sin} \left({2}^{k} x\right) = \cot \left({2}^{k - 1} x\right) - \cot \left({2}^{k} x\right)$

so we can build a telescopic series such that

( (1/sin(2x)=cot(x)-cot(2x)), (1/sin(2^2x)=cot(2x)-cot(2^2x)), (cdots=cdots), (1/sin(2^kx)=cot(2^(k-1)x)-cot(2^kx)) )

summing up we get

${\sum}_{k = 1}^{n} \frac{1}{\sin} \left({2}^{k} x\right) = \cot \left(x\right) - \cot \left({2}^{n} x\right)$