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sin3^n/3^n

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  • Sum of series:
  • lnn/n lnn/n
  • pi^(-n)/pi^n pi^(-n)/pi^n
  • √3 √3
  • sin3^n/3^n sin3^n/3^n

  • Identical expressions

  • sin three ^n/3^n
  • sinus of 3 to the power of n divide by 3 to the power of n
  • sinus of three to the power of n divide by 3 to the power of n
  • sin3n/3n
  • sin3^n divide by 3^n
Sum of series/ sin3^n/3^n

Sum of series sin3^n/3^n


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \       n   
  \   sin (3)
   )  -------
  /       n  
 /       3   
/___,        
n = 1
∑n=1∞sin⁡n(3)3n\sum_{n=1}^{\infty} \frac{\sin^{n}{\left(3 \right)}}{3^{n}} 
Sum(sin(3)^n/3^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
sin⁡n(3)3n\frac{\sin^{n}{\left(3 \right)}}{3^{n}}
It is a series of species
an(cx−x0)dna_{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:
Rd=x0+lim⁡n→∞∣anan+1∣cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=sin⁡n(3)a_{n} = \sin^{n}{\left(3 \right)}
and
x0=−3x_{0} = -3
,
d=−1d = -1
,
c=0c = 0
then
1R=∞~(−3+lim⁡n→∞(sin⁡n(3)sin⁡−n−1(3)))\frac{1}{R} = \tilde{\infty} \left(-3 + \lim_{n \to \infty}\left(\sin^{n}{\left(3 \right)} \sin^{- n - 1}{\left(3 \right)}\right)\right)
Let's take the limit
we find
False

False

R=0R = 0
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.04500.0500
The answer [src] 
    sin(3)    
--------------
  /    sin(3)\
3*|1 - ------|
  \      3   /
sin⁡(3)3(1−sin⁡(3)3)\frac{\sin{\left(3 \right)}}{3 \left(1 - \frac{\sin{\left(3 \right)}}{3}\right)} 
sin(3)/(3*(1 - sin(3)/3))
Numerical answer [src] 
0.0493619908697526004342609771427
0.0493619908697526004342609771427
The graph
Sum of series sin3^n/3^n

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