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cos(2n)/(2^n)

  • How to use it?

  • Sum of series:
  • cos(2*n)/2^n cos(2*n)/2^n
  • (n-1)/(n(n+1)(n+2)) (n-1)/(n(n+1)(n+2))
  • xx xx
  • sin(n*x)

  • Identical expressions

  • cos(two n)/(2^n)
  • co sinus of e of (2n) divide by (2 to the power of n)
  • co sinus of e of (two n) divide by (2 to the power of n)
  • cos(2n)/(2n)
  • cos2n/2n
  • cos2n/2^n
  • cos(2n) divide by (2^n)
Sum of series/ cos(2n)/(2^n)

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


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The solution

You have entered [src] 
  oo          
____          
\   `         
 \    cos(2*n)
  \   --------
  /       n   
 /       2    
/___,         
n = 1
$$\sum_{n=1}^{\infty} \frac{\cos{\left(2 n \right)}}{2^{n}}$$ 
Sum(cos(2*n)/2^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\cos{\left(2 n \right)}}{2^{n}}$$
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
$$a_{n} = \cos{\left(2 n \right)}$$
and
$$x_{0} = -2$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} \left|{\frac{\cos{\left(2 n \right)}}{\cos{\left(2 n + 2 \right)}}}\right|\right)$$
Let's take the limit
we find
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} \left|{\frac{\cos{\left(2 n \right)}}{\cos{\left(2 n + 2 \right)}}}\right|\right)$$
$$R = 0 \left(-2 + \lim_{n \to \infty} \left|{\frac{\cos{\left(2 n \right)}}{\cos{\left(2 n + 2 \right)}}}\right|\right)^{-1}$$
The rate of convergence of the power series
The answer [src] 
  oo              
 ___              
 \  `             
  \    -n         
  /   2  *cos(2*n)
 /__,             
n = 1
$$\sum_{n=1}^{\infty} 2^{- n} \cos{\left(2 n \right)}$$ 
Sum(2^(-n)*cos(2*n), (n, 1, oo))
Numerical answer [src] 
-0.274929801038943632397576981589
-0.274929801038943632397576981589
The graph
Sum of series cos(2n)/(2^n)

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