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sin(n)/2^n

  • How to use it?

  • Sum of series:
  • (4*9^n-2^n)/18^n (4*9^n-2^n)/18^n
  • k^2 k^2
  • (k^3-k^2) (k^3-k^2)
  • sin(n)/2^n sin(n)/2^n

  • Identical expressions

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

Sum of series sin(n)/2^n


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The solution

You have entered [src] 
  oo        
____        
\   `       
 \    sin(n)
  \   ------
  /      n  
 /      2   
/___,       
n = 1
$$\sum_{n=1}^{\infty} \frac{\sin{\left(n \right)}}{2^{n}}$$ 
Sum(sin(n)/2^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\sin{\left(n \right)}}{2^{n}}$$
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
$$a_{n} = \sin{\left(n \right)}$$
and
$$x_{0} = -2$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} \left|{\frac{\sin{\left(n \right)}}{\sin{\left(n + 1 \right)}}}\right|\right)$$
Let's take the limit
we find
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} \left|{\frac{\sin{\left(n \right)}}{\sin{\left(n + 1 \right)}}}\right|\right)$$
$$R = 0 \left(-2 + \lim_{n \to \infty} \left|{\frac{\sin{\left(n \right)}}{\sin{\left(n + 1 \right)}}}\right|\right)^{-1}$$
The rate of convergence of the power series
The answer [src] 
  oo            
 ___            
 \  `           
  \    -n       
  /   2  *sin(n)
 /__,           
n = 1
$$\sum_{n=1}^{\infty} 2^{- n} \sin{\left(n \right)}$$ 
Sum(2^(-n)*sin(n), (n, 1, oo))
Numerical answer [src] 
0.592837620697942576555228456977
0.592837620697942576555228456977
The graph
Sum of series sin(n)/2^n

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