Prove that $n \sum k = 1 \frac{1}{{sin 2}^{k} x} = cot x - {cot 2}^{n} x$ $sum_(k=1)^n 1/(sin2^kx)=cot x - cot 2^nx$ for every $x ne (kpi)/2^k, x in RR, n in NN^+$ ?

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Answer 1 · 2016-12-01T15:18:22

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

I get a similar result for

$sum 1/sin^(2^k)x$

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

using $1+X + X^2+...+X^n=(1-X^(n+1))/(1-X)$

$=-tan^2x(1-csc^(2^(n+1))x)-1$

$=sec^2x(csc^(2^n) x-1)$ .

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

Answer 2 · 2016-12-01T16:39:32

See below.

Always is worth to read Ramanujan's Third Notebook.

With the identity

$cot(x)=cot(x/2)-csc(x)$ we get

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

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

$sum_(k=1)^n 1/sin(2^kx) = cot(x)-cot(2^nx)$

Prove that sum_(k=1)^n 1/(sin2^kx)=cot x - cot 2^nxn∑k=11sin2kx=cotx−cot2nxsum_(k=1)^n 1/(sin2^kx)=cot x - cot 2^nx for every x ne (kpi)/2^k, x in RR, n in NN^+x ne (kpi)/2^k, x in RR, n in NN^+?