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sin(n)^2/n^(3/2)

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  • Sum of series:
  • sin(n)^2/n^(3/2) sin(n)^2/n^(3/2)
  • e^ipi/n/n
  • 13n(13n+13)-(13n-13)(13+13n) 13n(13n+13)-(13n-13)(13+13n)
  • n*x^(6n)

  • Identical expressions

  • sin(n)^ two /n^(three / two)
  • sinus of (n) squared divide by n to the power of (3 divide by 2)
  • sinus of (n) to the power of two divide by n to the power of (three divide by two)
  • sin(n)2/n(3/2)
  • sinn2/n3/2
  • sin(n)²/n^(3/2)
  • sin(n) to the power of 2/n to the power of (3/2)
  • sinn^2/n^3/2
  • sin(n)^2 divide by n^(3 divide by 2)
Sum of series/ sin(n)^2/n^(3/2)

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


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The solution

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

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