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  • Sum of series:
  • sin(pi/2^n)
  • 1/(n*lnn) 1/(n*lnn)
  • (lnx-1)/(ln(x+1)-1)
  • n*x^n

  • Identical expressions

  • sin(pi/ two ^n)
  • sinus of ( Pi divide by 2 to the power of n)
  • sinus of ( Pi divide by two to the power of n)
  • sin(pi/2n)
  • sinpi/2n
  • sinpi/2^n
  • sin(pi divide by 2^n)
Sum of series/ sin(pi/2^n)

Sum of series sin(pi/2^n)


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \       /pi\
  \   sin|--|
  /      | n|
 /       \2 /
/___,        
n = 1
$$\sum_{n=1}^{\infty} \sin{\left(\frac{\pi}{2^{n}} \right)}$$ 
Sum(sin(pi/2^n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\sin{\left(\frac{\pi}{2^{n}} \right)}$$
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
$$a_{n} = \sin{\left(2^{- n} \pi \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2^{- n} \pi \right)}}{\sin{\left(2^{- n - 1} \pi \right)}}}\right|$$
Let's take the limit
we find
False

False
The answer [src] 
  oo             
 ___             
 \  `            
  \      /    -n\
  /   sin\pi*2  /
 /__,            
n = 1
$$\sum_{n=1}^{\infty} \sin{\left(2^{- n} \pi \right)}$$ 
Sum(sin(pi*2^(-n)), (n, 1, oo))
Numerical answer [src] 
2.48104991933372426275028794950
2.48104991933372426275028794950

    Examples of finding the sum of a series

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