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  • Sum of series:
  • 1/(n*(n+3)) 1/(n*(n+3))
  • (13^(2n-1))/((2^n)*(n-1)!) (13^(2n-1))/((2^n)*(n-1)!)
  • e^n/n^10 e^n/n^10
  • (n^2+5)/2^n (n^2+5)/2^n

  • Identical expressions

  • sin^ two (x)/(x^(one / three))
  • sinus of squared (x) divide by (x to the power of (1 divide by 3))
  • sinus of to the power of two (x) divide by (x to the power of (one divide by three))
  • sin2(x)/(x(1/3))
  • sin2x/x1/3
  • sin²(x)/(x^(1/3))
  • sin to the power of 2(x)/(x to the power of (1/3))
  • sin^2x/x^1/3
  • sin^2(x) divide by (x^(1 divide by 3))
Sum of series/ sin^2(x)/(x^(1/3))

Sum of series sin^2(x)/(x^(1/3))


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \       2   
  \   sin (x)
   )  -------
  /    3 ___ 
 /     \/ x  
/___,        
n = 1
∑n=1∞sin⁡2(x)x3\sum_{n=1}^{\infty} \frac{\sin^{2}{\left(x \right)}}{\sqrt[3]{x}} 
Sum(sin(x)^2/x^(1/3), (n, 1, oo))
The radius of convergence of the power series
Given number:
sin⁡2(x)x3\frac{\sin^{2}{\left(x \right)}}{\sqrt[3]{x}}
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⁡2(x)x3a_{n} = \frac{\sin^{2}{\left(x \right)}}{\sqrt[3]{x}}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞11 = \lim_{n \to \infty} 1
Let's take the limit
we find
True

False
The answer [src] 
      2   
oo*sin (x)
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  3 ___   
  \/ x
∞sin⁡2(x)x3\frac{\infty \sin^{2}{\left(x \right)}}{\sqrt[3]{x}} 
oo*sin(x)^2/x^(1/3)

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