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arctg1/(2n^2)

  • How to use it?

  • Sum of series:
  • (3^n-4^n)/12^n (3^n-4^n)/12^n
  • sin(n*x)/n^2
  • arctg1/(2n^2) arctg1/(2n^2)
  • 28 28

  • Identical expressions

  • arctg1/(two n^2)
  • arctg1 divide by (2n squared )
  • arctg1 divide by (two n squared )
  • arctg1/(2n2)
  • arctg1/2n2
  • arctg1/(2n²)
  • arctg1/(2n to the power of 2)
  • arctg1/2n^2
  • arctg1 divide by (2n^2)
Sum of series/ arctg1/(2n^2)

Sum of series arctg1/(2n^2)


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \    atan(1)
  \   -------
  /        2 
 /      2*n  
/___,        
n = 1
∑n=1∞atan⁡(1)2n2\sum_{n=1}^{\infty} \frac{\operatorname{atan}{\left(1 \right)}}{2 n^{2}} 
Sum(atan(1)/((2*n^2)), (n, 1, oo))
The radius of convergence of the power series
Given number:
atan⁡(1)2n2\frac{\operatorname{atan}{\left(1 \right)}}{2 n^{2}}
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=π8n2a_{n} = \frac{\pi}{8 n^{2}}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞((n+1)2n2)1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.20.6
The answer [src] 
  3
pi 
---
 48
π348\frac{\pi^{3}}{48} 
pi^3/48
Numerical answer [src] 
0.645964097506246253655756563898
0.645964097506246253655756563898
The graph
Sum of series arctg1/(2n^2)

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