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arctg(1/(2*n^2))

  • How to use it?

  • Sum of series:
  • n^(2/3)*arctg(1/n^2) n^(2/3)*arctg(1/n^2)
  • 1/4^1 1/4^1
  • 2n 2n
  • 2/(n^2+6n+8) 2/(n^2+6n+8)

  • Identical expressions

  • arctg(one /(two *n^ two))
  • arctg(1 divide by (2 multiply by n squared ))
  • arctg(one divide by (two multiply by n to the power of two))
  • arctg(1/(2*n2))
  • arctg1/2*n2
  • arctg(1/(2*n²))
  • arctg(1/(2*n to the power of 2))
  • arctg(1/(2n^2))
  • arctg(1/(2n2))
  • arctg1/2n2
  • arctg1/2n^2
  • arctg(1 divide by (2*n^2))
Sum of series/ arctg(1/(2*n^2))

Sum of series arctg(1/(2*n^2))


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The solution

You have entered [src] 
  oo            
____            
\   `           
 \        / 1  \
  \   atan|----|
  /       |   2|
 /        \2*n /
/___,           
n = 1
∑n=1∞atan⁡(12n2)\sum_{n=1}^{\infty} \operatorname{atan}{\left(\frac{1}{2 n^{2}} \right)} 
Sum(atan(1/(2*n^2)), (n, 1, oo))
The radius of convergence of the power series
Given number:
atan⁡(12n2)\operatorname{atan}{\left(\frac{1}{2 n^{2}} \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=atan⁡(12n2)a_{n} = \operatorname{atan}{\left(\frac{1}{2 n^{2}} \right)}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞(atan⁡(12n2)atan⁡(12(n+1)2))1 = \lim_{n \to \infty}\left(\frac{\operatorname{atan}{\left(\frac{1}{2 n^{2}} \right)}}{\operatorname{atan}{\left(\frac{1}{2 \left(n + 1\right)^{2}} \right)}}\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.250.75
Numerical answer [src] 
0.785398163397448309615660845820
0.785398163397448309615660845820
The graph
Sum of series arctg(1/(2*n^2))

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