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factorial(n)/(n^3+n+8)

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  • Sum of series:
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  • factorial(n)/(n^3+n+8) factorial(n)/(n^3+n+8)
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  • 0.1 0.1

  • Identical expressions

  • factorial(n)/(n^ three +n+ eight)
  • factorial(n) divide by (n cubed plus n plus 8)
  • factorial(n) divide by (n to the power of three plus n plus eight)
  • factorial(n)/(n3+n+8)
  • factorialn/n3+n+8
  • factorial(n)/(n³+n+8)
  • factorial(n)/(n to the power of 3+n+8)
  • factorialn/n^3+n+8
  • factorial(n) divide by (n^3+n+8)

  • Similar expressions

  • factorial(n)/(n^3-n+8)
  • factorial(n)/(n^3+n-8)
Sum of series/ factorial(n)/(n^3+n+8)

Sum of series factorial(n)/(n^3+n+8)


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The solution

You have entered [src] 
  oo            
____            
\   `           
 \        n!    
  \   ----------
  /    3        
 /    n  + n + 8
/___,           
n = 1
∑n=1∞n!(n3+n)+8\sum_{n=1}^{\infty} \frac{n!}{\left(n^{3} + n\right) + 8} 
Sum(factorial(n)/(n^3 + n + 8), (n, 1, oo))
The radius of convergence of the power series
Given number:
n!(n3+n)+8\frac{n!}{\left(n^{3} + 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=n!n3+n+8a_{n} = \frac{n!}{n^{3} + n + 8}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞((n+(n+1)3+9)∣n!(n+1)!∣n3+n+8)1 = \lim_{n \to \infty}\left(\frac{\left(n + \left(n + 1\right)^{3} + 9\right) \left|{\frac{n!}{\left(n + 1\right)!}}\right|}{n^{3} + n + 8}\right)
Let's take the limit
we find
False

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.5020
Numerical answer
The series diverges
The graph
Sum of series factorial(n)/(n^3+n+8)

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