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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
  • ln(n+1)/n ln(n+1)/n
  • log(1+1/n)-log(1+1/(n+1)) log(1+1/n)-log(1+1/(n+1))

  • Identical expressions

  • sin^ two (x)/ three ^n
  • sinus of squared (x) divide by 3 to the power of n
  • sinus of to the power of two (x) divide by three to the power of n
  • sin2(x)/3n
  • sin2x/3n
  • sin²(x)/3^n
  • sin to the power of 2(x)/3 to the power of n
  • sin^2x/3^n
  • sin^2(x) divide by 3^n
Sum of series/ sin^2(x)/3^n

Sum of series sin^2(x)/3^n


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \       2   
  \   sin (x)
   )  -------
  /       n  
 /       3   
/___,        
n = 1
$$\sum_{n=1}^{\infty} \frac{\sin^{2}{\left(x \right)}}{3^{n}}$$ 
Sum(sin(x)^2/3^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\sin^{2}{\left(x \right)}}{3^{n}}$$
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
$$a_{n} = \sin^{2}{\left(x \right)}$$
and
$$x_{0} = -3$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-3 + \lim_{n \to \infty} 1\right)$$
Let's take the limit
we find
False

$$R = 0$$
The answer [src] 
   2   
sin (x)
-------
   2
$$\frac{\sin^{2}{\left(x \right)}}{2}$$ 
sin(x)^2/2

    Examples of finding the sum of a series

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