Find the sum of the series (1/2^n) tan (x/2^n) from n=1 to infinity for every value of x?

Question

Find the sum of the series from n=1 to infinity: [1/(2^n)] tan[x/(2^n)] for every value of x.

Answer 1

I don't know, but I have found that:
for x = pi/2, the sum is 2/pi
and for x = pi/4 the sum is (4/pi) - 1
and for x = pi/8 the sum is (8/pi) - (1 + sqrt(2))
and for x = pi/3 the sum is (3/pi) - 3/sqrt(3)
and for x = pi/6 the sum is (6/pi) - sqrt(3)
I found these by doing partial sums for several values of x, then looking for numbers I recognize. This led to the first two finds, then the rest were found by matching the expected pattern (i.e. sum(x) = 1/x - y(x), solve for y(x) and try to recognize it). I don't recognize any other partial sums, but I also found by graphing that this is an odd function on (-pi,pi).
That may help you along.

I am assuming you mean (1/(2^n)) tan (x/(2^n))


New: I got it! sum(x) = (1/x) - cot(x)

Now it makes sense!