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  • Sum of series:
  • 5/n 5/n
  • (n-2^(3*n))/(n+1)^(1/3) (n-2^(3*n))/(n+1)^(1/3)
  • ((3*x)^n)/n!
  • 1/((5n-4)(5n+1)) 1/((5n-4)(5n+1))

  • Identical expressions

  • ((three *x)^n)/n!
  • ((3 multiply by x) to the power of n) divide by n!
  • ((three multiply by x) to the power of n) divide by n!
  • ((3*x)n)/n!
  • 3*xn/n!
  • ((3x)^n)/n!
  • ((3x)n)/n!
  • 3xn/n!
  • 3x^n/n!
  • ((3*x)^n) divide by n!
Sum of series/ ((3*x)^n)/n!

Sum of series ((3*x)^n)/n!


∑
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The solution

You have entered [src] 
  oo        
____        
\   `       
 \         n
  \   (3*x) 
  /   ------
 /      n!  
/___,       
n = 2
∑n=2∞(3x)nn!\sum_{n=2}^{\infty} \frac{\left(3 x\right)^{n}}{n!} 
Sum((3*x)^n/factorial(n), (n, 2, oo))
The radius of convergence of the power series
Given number:
(3x)nn!\frac{\left(3 x\right)^{n}}{n!}
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=1n!a_{n} = \frac{1}{n!}
and
x0=0x_{0} = 0
,
d=1d = 1
,
c=3c = 3
then
R=lim⁡n→∞∣(n+1)!n!∣3R = \frac{\lim_{n \to \infty} \left|{\frac{\left(n + 1\right)!}{n!}}\right|}{3}
Let's take the limit
we find
R=∞R = \infty
The answer [src] 
     /  2   2*x         \
     |- - - ---      3*x|
   2 |  9    3    2*e   |
9*x *|--------- + ------|
     |     2          2 |
     \    x        9*x  /
-------------------------
            2
9x2(−2x3−29x2+2e3x9x2)2\frac{9 x^{2} \left(\frac{- \frac{2 x}{3} - \frac{2}{9}}{x^{2}} + \frac{2 e^{3 x}}{9 x^{2}}\right)}{2} 
9*x^2*((-2/9 - 2*x/3)/x^2 + 2*exp(3*x)/(9*x^2))/2

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