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sin(n^n*factorial(n))/2^n
  • How to use it?

  • Sum of series:
  • x^n/2^n
  • 1/((3n-2)*(3n+1)) 1/((3n-2)*(3n+1))
  • (-1)^n/2n-1 (-1)^n/2n-1
  • (x-1)^n/5^n
  • Identical expressions

  • sin(n^n*factorial(n))/ two ^n
  • sinus of (n to the power of n multiply by factorial(n)) divide by 2 to the power of n
  • sinus of (n to the power of n multiply by factorial(n)) divide by two to the power of n
  • sin(nn*factorial(n))/2n
  • sinnn*factorialn/2n
  • sin(n^nfactorial(n))/2^n
  • sin(nnfactorial(n))/2n
  • sinnnfactorialn/2n
  • sinn^nfactorialn/2^n
  • sin(n^n*factorial(n)) divide by 2^n

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



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The solution

You have entered [src]
  oo            
____            
\   `           
 \       / n   \
  \   sin\n *n!/
   )  ----------
  /        n    
 /        2     
/___,           
n = 1           
$$\sum_{n=1}^{\infty} \frac{\sin{\left(n^{n} n! \right)}}{2^{n}}$$
Sum(sin(n^n*factorial(n))/2^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\sin{\left(n^{n} n! \right)}}{2^{n}}$$
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
$$a_{n} = \sin{\left(n^{n} n! \right)}$$
and
$$x_{0} = -2$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} \left|{\frac{\sin{\left(n^{n} n! \right)}}{\sin{\left(\left(n + 1\right)^{n + 1} \left(n + 1\right)! \right)}}}\right|\right)$$
Let's take the limit
we find
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} \left|{\frac{\sin{\left(n^{n} n! \right)}}{\sin{\left(\left(n + 1\right)^{n + 1} \left(n + 1\right)! \right)}}}\right|\right)$$
$$R = 0 \left(-2 + \lim_{n \to \infty} \left|{\frac{\sin{\left(n^{n} n! \right)}}{\sin{\left(\left(n + 1\right)^{n + 1} \left(n + 1\right)! \right)}}}\right|\right)^{-1}$$
The rate of convergence of the power series
The answer [src]
  oo                
 ___                
 \  `               
  \    -n    / n   \
  /   2  *sin\n *n!/
 /__,               
n = 1               
$$\sum_{n=1}^{\infty} 2^{- n} \sin{\left(n^{n} n! \right)}$$
Sum(2^(-n)*sin(n^n*factorial(n)), (n, 1, oo))
Numerical answer [src]
0.506568681111421253210866019752
0.506568681111421253210866019752
The graph
Sum of series sin(n^n*factorial(n))/2^n

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