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factorial(n)/(n^3+n^2+4)

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  • Sum of series:
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  • (2n-1)/2^n (2n-1)/2^n
  • sin(3n)/(7n)^(1/5) sin(3n)/(7n)^(1/5)
  • arctg1/(2n^2) arctg1/(2n^2)

  • Identical expressions

  • factorial(n)/(n^ three +n^ two + four)
  • factorial(n) divide by (n cubed plus n squared plus 4)
  • factorial(n) divide by (n to the power of three plus n to the power of two plus four)
  • factorial(n)/(n3+n2+4)
  • factorialn/n3+n2+4
  • factorial(n)/(n³+n²+4)
  • factorial(n)/(n to the power of 3+n to the power of 2+4)
  • factorialn/n^3+n^2+4
  • factorial(n) divide by (n^3+n^2+4)

  • Similar expressions

  • factorial(n)/(n^3+n^2-4)
  • factorial(n)/(n^3-n^2+4)
Sum of series/ factorial(n)/(n^3+n^2+4)

Sum of series factorial(n)/(n^3+n^2+4)


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The solution

You have entered [src] 
  oo             
____             
\   `            
 \         n!    
  \   -----------
  /    3    2    
 /    n  + n  + 4
/___,            
n = 1
$$\sum_{n=1}^{\infty} \frac{n!}{\left(n^{3} + n^{2}\right) + 4}$$ 
Sum(factorial(n)/(n^3 + n^2 + 4), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{n!}{\left(n^{3} + n^{2}\right) + 4}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \frac{n!}{n^{3} + n^{2} + 4}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(\left(n + 1\right)^{3} + \left(n + 1\right)^{2} + 4\right) \left|{\frac{n!}{\left(n + 1\right)!}}\right|}{n^{3} + n^{2} + 4}\right)$$
Let's take the limit
we find
False

False
The rate of convergence of the power series
Numerical answer
The series diverges
The graph
Sum of series factorial(n)/(n^3+n^2+4)

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