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  • Sum of series:
  • sin(n*x)/n^3
  • 1/4n^2-1 1/4n^2-1
  • ((8)(ln(x)))/(x^3)
  • cos(npi*2)/n^4 cos(npi*2)/n^4

  • Identical expressions

  • ((eight)(ln(x)))/(x^ three)
  • ((8)(ln(x))) divide by (x cubed )
  • ((eight)(ln(x))) divide by (x to the power of three)
  • ((8)(ln(x)))/(x3)
  • 8lnx/x3
  • ((8)(ln(x)))/(x³)
  • ((8)(ln(x)))/(x to the power of 3)
  • 8lnx/x^3
  • ((8)(ln(x))) divide by (x^3)
Sum of series/ ((8)(ln(x)))/(x^3)

Sum of series ((8)(ln(x)))/(x^3)


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The solution

You have entered [src] 
  oo          
____          
\   `         
 \    8*log(x)
  \   --------
  /       3   
 /       x    
/___,         
n = 1
$$\sum_{n=1}^{\infty} \frac{8 \log{\left(x \right)}}{x^{3}}$$ 
Sum((8*log(x))/x^3, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{8 \log{\left(x \right)}}{x^{3}}$$
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
$$a_{n} = \frac{8 \log{\left(x \right)}}{x^{3}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find
True

False
The answer [src] 
oo*log(x)
---------
     3   
    x
$$\frac{\infty \log{\left(x \right)}}{x^{3}}$$ 
oo*log(x)/x^3

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