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^+#?

Question

3913 views around the world

You can reuse this answer
Creative Commons License

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)#