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sin(pi/2^(n-1))

Sum of series sin(pi/2^(n-1))



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The solution

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  oo             
____             
\   `            
 \       /  pi  \
  \   sin|------|
  /      | n - 1|
 /       \2     /
/___,            
n = 1            
$$\sum_{n=1}^{\infty} \sin{\left(\frac{\pi}{2^{n - 1}} \right)}$$
Sum(sin(pi/2^(n - 1)), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\sin{\left(\frac{\pi}{2^{n - 1}} \right)}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \sin{\left(2^{1 - n} \pi \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2^{1 - n} \pi \right)}}{\sin{\left(2^{- n} \pi \right)}}}\right|$$
Let's take the limit
we find
False

False
The rate of convergence of the power series
The answer [src]
0
$$0$$
0
Numerical answer [src]
2.48104991933372426275028794950
2.48104991933372426275028794950
The graph
Sum of series sin(pi/2^(n-1))

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