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  • Sum of series:
  • (((-1)^(n))*(16*x)^(n))/n!
  • (3×x^n)/(4n^2-9)
  • x(1-x)^n
  • log(z^4) log(z^4)

  • Identical expressions

  • four !^ two (i- one)/n^ four
  • 4! squared (i minus 1) divide by n to the power of 4
  • four ! to the power of two (i minus one) divide by n to the power of four
  • 4!2(i-1)/n4
  • 4!2i-1/n4
  • 4!²(i-1)/n⁴
  • 4! to the power of 2(i-1)/n to the power of 4
  • 4!^2i-1/n^4
  • 4!^2(i-1) divide by n^4

  • Similar expressions

  • 4!^2(i+1)/n^4
Sum of series/ 4!^2(i-1)/n^4

Sum of series 4!^2(i-1)/n^4


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The solution

You have entered [src] 
  oo             
____             
\   `            
 \      2        
  \   4! *(i - 1)
   )  -----------
  /         4    
 /         n     
/___,            
i = 1
$$\sum_{i=1}^{\infty} \frac{\left(i - 1\right) 4!^{2}}{n^{4}}$$ 
Sum((factorial(4)^2*(i - 1))/n^4, (i, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(i - 1\right) 4!^{2}}{n^{4}}$$
It is a series of species
$$a_{i} \left(c x - x_{0}\right)^{d i}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{i \to \infty} \left|{\frac{a_{i}}{a_{i + 1}}}\right|}{c}$$
In this case
$$a_{i} = \frac{576 i - 576}{n^{4}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{i \to \infty}\left(\frac{\left|{576 i - 576}\right|}{576 i}\right)$$
Let's take the limit
we find
True

False
The answer [src] 
oo
--
 4
n
$$\frac{\infty}{n^{4}}$$ 
oo/n^4

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