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sin^2(1/4n)

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  • Sum of series:
  • factorial(n)*3^n/n^n factorial(n)*3^n/n^n
  • sin^2(1/4n) sin^2(1/4n)
  • -3 -3
  • 17 17

  • Identical expressions

  • sin^ two (one /4n)
  • sinus of squared (1 divide by 4n)
  • sinus of to the power of two (one divide by 4n)
  • sin2(1/4n)
  • sin21/4n
  • sin²(1/4n)
  • sin to the power of 2(1/4n)
  • sin^21/4n
  • sin^2(1 divide by 4n)
Sum of series/ sin^2(1/4n)

Sum of series sin^2(1/4n)


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The solution

You have entered [src] 
  oo         
 ___         
 \  `        
  \      2/n\
   )  sin |-|
  /       \4/
 /__,        
n = 1
∑n=1∞sin⁡2(n4)\sum_{n=1}^{\infty} \sin^{2}{\left(\frac{n}{4} \right)} 
Sum(sin(n/4)^2, (n, 1, oo))
The radius of convergence of the power series
Given number:
sin⁡2(n4)\sin^{2}{\left(\frac{n}{4} \right)}
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(n4)a_{n} = \sin^{2}{\left(\frac{n}{4} \right)}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞(sin⁡2(n4)∣1sin⁡2(n4+14)∣)1 = \lim_{n \to \infty}\left(\sin^{2}{\left(\frac{n}{4} \right)} \left|{\frac{1}{\sin^{2}{\left(\frac{n}{4} + \frac{1}{4} \right)}}}\right|\right)
Let's take the limit
we find
1=lim⁡n→∞(sin⁡2(n4)∣1sin⁡2(n4+14)∣)1 = \lim_{n \to \infty}\left(\sin^{2}{\left(\frac{n}{4} \right)} \left|{\frac{1}{\sin^{2}{\left(\frac{n}{4} + \frac{1}{4} \right)}}}\right|\right)
False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.505
Numerical answer
The series diverges
The graph
Sum of series sin^2(1/4n)

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