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n^100*100^n/factorial(n)

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  • Sum of series:
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  • 1/3^n 1/3^n
  • sin(1/n) sin(1/n)
  • a^n/factorial(n)

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  • n^ one hundred * one hundred ^n/factorial(n)
  • n to the power of 100 multiply by 100 to the power of n divide by factorial(n)
  • n to the power of one hundred multiply by one hundred to the power of n divide by factorial(n)
  • n100*100n/factorial(n)
  • n100*100n/factorialn
  • n^100100^n/factorial(n)
  • n100100n/factorial(n)
  • n100100n/factorialn
  • n^100100^n/factorialn
  • n^100*100^n divide by factorial(n)
Sum of series/ n^100*100^n/factorial(n)

Sum of series n^100*100^n/factorial(n)


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The solution

You have entered [src] 
  oo           
____           
\   `          
 \     100    n
  \   n   *100 
  /   ---------
 /        n!   
/___,          
n = 1
$$\sum_{n=1}^{\infty} \frac{100^{n} n^{100}}{n!}$$ 
Sum((n^100*100^n)/factorial(n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{100^{n} n^{100}}{n!}$$
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
$$a_{n} = \frac{n^{100}}{n!}$$
and
$$x_{0} = -100$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(-100 + \lim_{n \to \infty}\left(\frac{n^{100} \left|{\frac{\left(n + 1\right)!}{n!}}\right|}{\left(n + 1\right)^{100}}\right)\right)$$
Let's take the limit
we find
$$R = \infty$$
The rate of convergence of the power series
Numerical answer [src] 
0.e+257
0.e+257
The graph
Sum of series n^100*100^n/factorial(n)

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