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n(1-cos(pi/n^2))

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  • Sum of series:
  • 1/(2^n) 1/(2^n)
  • (1+n)/(1+n^2) (1+n)/(1+n^2)
  • 1/4n^2+1 1/4n^2+1
  • cos(n)*x/n^2

  • Identical expressions

  • n(one -cos(pi/n^ two))
  • n(1 minus co sinus of e of ( Pi divide by n squared ))
  • n(one minus co sinus of e of ( Pi divide by n to the power of two))
  • n(1-cos(pi/n2))
  • n1-cospi/n2
  • n(1-cos(pi/n²))
  • n(1-cos(pi/n to the power of 2))
  • n1-cospi/n^2
  • n(1-cos(pi divide by n^2))

  • Similar expressions

  • n(1+cos(pi/n^2))
Sum of series/ n(1-cos(pi/n^2))

Sum of series n(1-cos(pi/n^2))


∑
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The solution

You have entered [src] 
  oo                 
____                 
\   `                
 \      /       /pi\\
  \   n*|1 - cos|--||
  /     |       | 2||
 /      \       \n //
/___,                
n = 1
∑n=1∞n(1−cos⁡(πn2))\sum_{n=1}^{\infty} n \left(1 - \cos{\left(\frac{\pi}{n^{2}} \right)}\right) 
Sum(n*(1 - cos(pi/n^2)), (n, 1, oo))
The radius of convergence of the power series
Given number:
n(1−cos⁡(πn2))n \left(1 - \cos{\left(\frac{\pi}{n^{2}} \right)}\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=n(1−cos⁡(πn2))a_{n} = n \left(1 - \cos{\left(\frac{\pi}{n^{2}} \right)}\right)
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞(n∣cos⁡(πn2)−1cos⁡(π(n+1)2)−1∣n+1)1 = \lim_{n \to \infty}\left(\frac{n \left|{\frac{\cos{\left(\frac{\pi}{n^{2}} \right)} - 1}{\cos{\left(\frac{\pi}{\left(n + 1\right)^{2}} \right)} - 1}}\right|}{n + 1}\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.513
Numerical answer [src] 
2.96387625930109511160298699125
2.96387625930109511160298699125
The graph
Sum of series n(1-cos(pi/n^2))

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