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9^n*factorial(n)/n^(n*2)

  • How to use it?

  • Sum of series:
  • 1+1/factorial(n) 1+1/factorial(n)
  • (n+1)^n/(n^(n+2)) (n+1)^n/(n^(n+2))
  • arctg(1/(n^2+n+1)) arctg(1/(n^2+n+1))
  • 9^n*factorial(n)/n^(n*2) 9^n*factorial(n)/n^(n*2)

  • Identical expressions

  • nine ^n*factorial(n)/n^(n* two)
  • 9 to the power of n multiply by factorial(n) divide by n to the power of (n multiply by 2)
  • nine to the power of n multiply by factorial(n) divide by n to the power of (n multiply by two)
  • 9n*factorial(n)/n(n*2)
  • 9n*factorialn/nn*2
  • 9^nfactorial(n)/n^(n2)
  • 9nfactorial(n)/n(n2)
  • 9nfactorialn/nn2
  • 9^nfactorialn/n^n2
  • 9^n*factorial(n) divide by n^(n*2)
Sum of series/ 9^n*factorial(n)/n^(n*2)

Sum of series 9^n*factorial(n)/n^(n*2)


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The solution

You have entered [src] 
  oo       
____       
\   `      
 \     n   
  \   9 *n!
   )  -----
  /     n*2
 /     n   
/___,      
n = 1
$$\sum_{n=1}^{\infty} \frac{9^{n} n!}{n^{2 n}}$$ 
Sum((9^n*factorial(n))/n^(n*2), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{9^{n} n!}{n^{2 n}}$$
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
$$a_{n} = n^{- 2 n} n!$$
and
$$x_{0} = -9$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(-9 + \lim_{n \to \infty}\left(n^{- 2 n} \left(n + 1\right)^{2 n + 2} \left|{\frac{n!}{\left(n + 1\right)!}}\right|\right)\right)$$
Let's take the limit
we find
$$R = \infty$$
The rate of convergence of the power series
The answer [src] 
  oo             
 ___             
 \  `            
  \    n  -2*n   
  /   9 *n    *n!
 /__,            
n = 1
$$\sum_{n=1}^{\infty} 9^{n} n^{- 2 n} n!$$ 
Sum(9^n*n^(-2*n)*factorial(n), (n, 1, oo))
Numerical answer [src] 
28.4718751531196540595858172056
28.4718751531196540595858172056
The graph
Sum of series 9^n*factorial(n)/n^(n*2)

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