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cos(i*n)/2^n

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  • Sum of series:
  • (n+1)^2/2^(n-1) (n+1)^2/2^(n-1)
  • (-1)^n/n (-1)^n/n
  • cos(i*n)/2^n cos(i*n)/2^n
  • sin1/n sin1/n

  • Identical expressions

  • cos(i*n)/ two ^n
  • co sinus of e of (i multiply by n) divide by 2 to the power of n
  • co sinus of e of (i multiply by n) divide by two to the power of n
  • cos(i*n)/2n
  • cosi*n/2n
  • cos(in)/2^n
  • cos(in)/2n
  • cosin/2n
  • cosin/2^n
  • cos(i*n) divide by 2^n
Sum of series/ cos(i*n)/2^n

Sum of series cos(i*n)/2^n


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The solution

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

False

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
  oo             
 ___             
 \  `            
  \    -n        
  /   2  *cosh(n)
 /__,            
n = 1
$$\sum_{n=1}^{\infty} 2^{- n} \cosh{\left(n \right)}$$ 
Sum(2^(-n)*cosh(n), (n, 1, oo))
The graph
Sum of series cos(i*n)/2^n

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