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

  • How to use it?

  • Sum of series:
  • 2i 2i
  • lnn/n lnn/n
  • factorial(n+2)/n^n factorial(n+2)/n^n
  • sin(pi/2^(n-1)) sin(pi/2^(n-1))

  • Identical expressions

  • sin(pi/ two ^(n- one))
  • sinus of ( Pi divide by 2 to the power of (n minus 1))
  • sinus of ( Pi divide by two to the power of (n minus one))
  • sin(pi/2(n-1))
  • sinpi/2n-1
  • sinpi/2^n-1
  • sin(pi divide by 2^(n-1))

  • Similar expressions

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

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


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The solution

You have entered [src] 
  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))

    Examples of finding the sum of a series

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