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1/(n*ln^2n)

  • How to use it?

  • Sum of series:
  • 1/(n*ln^2n) 1/(n*ln^2n)
  • factorial(n)/(n+1) factorial(n)/(n+1)
  • 5/((5n-2)*(5n+3)) 5/((5n-2)*(5n+3))
  • sin(pi/n)^n sin(pi/n)^n

  • Identical expressions

  • one /(n*ln^2n)
  • 1 divide by (n multiply by ln squared n)
  • one divide by (n multiply by ln squared n)
  • 1/(n*ln2n)
  • 1/n*ln2n
  • 1/(n*ln²n)
  • 1/(n*ln to the power of 2n)
  • 1/(nln^2n)
  • 1/(nln2n)
  • 1/nln2n
  • 1/nln^2n
  • 1 divide by (n*ln^2n)
Sum of series/ 1/(n*ln^2n)

Sum of series 1/(n*ln^2n)


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The solution

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

False
The rate of convergence of the power series
The graph
Sum of series 1/(n*ln^2n)

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