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2/(n^2+6n+8)

  • 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

  • two /(n^ two +6n+ eight)
  • 2 divide by (n squared plus 6n plus 8)
  • two divide by (n to the power of two plus 6n plus eight)
  • 2/(n2+6n+8)
  • 2/n2+6n+8
  • 2/(n²+6n+8)
  • 2/(n to the power of 2+6n+8)
  • 2/n^2+6n+8
  • 2 divide by (n^2+6n+8)

  • Similar expressions

  • 2/(n^2+6n-8)
  • 2/(n^2-6n+8)
Sum of series/ 2/(n^2+6n+8)

Sum of series 2/(n^2+6n+8)


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The solution

You have entered [src] 
  oo              
____              
\   `             
 \         2      
  \   ------------
  /    2          
 /    n  + 6*n + 8
/___,             
n = 1
∑n=1∞2(n2+6n)+8\sum_{n=1}^{\infty} \frac{2}{\left(n^{2} + 6 n\right) + 8} 
Sum(2/(n^2 + 6*n + 8), (n, 1, oo))
The radius of convergence of the power series
Given number:
2(n2+6n)+8\frac{2}{\left(n^{2} + 6 n\right) + 8}
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=2n2+6n+8a_{n} = \frac{2}{n^{2} + 6 n + 8}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞(2(3n+(n+1)22+7)n2+6n+8)1 = \lim_{n \to \infty}\left(\frac{2 \left(3 n + \frac{\left(n + 1\right)^{2}}{2} + 7\right)}{n^{2} + 6 n + 8}\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.000.50
The answer [src] 
7/12
712\frac{7}{12} 
7/12
Numerical answer [src] 
0.583333333333333333333333333333
0.583333333333333333333333333333
The graph
Sum of series 2/(n^2+6n+8)

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