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  • Sum of series:
  • 1/(3n-2)*(3n+1) 1/(3n-2)*(3n+1)
  • x^n/sqrt(n+1)
  • 1/ln(n) 1/ln(n)
  • 2n 2n

  • Identical expressions

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

Sum of series (sin*(pi/2^n))^n


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The solution

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

False
The answer [src] 
  oo              
 ___              
 \  `             
  \      n/    -n\
  /   sin \pi*2  /
 /__,             
n = 1
$$\sum_{n=1}^{\infty} \sin^{n}{\left(2^{- n} \pi \right)}$$ 
Sum(sin(pi*2^(-n))^n, (n, 1, oo))
Numerical answer [src] 
1.55750033361599519690376036153
1.55750033361599519690376036153

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