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1/(n((lnn)^2))

  • How to use it?

  • Sum of series:
  • sin(kx/n)
  • 1/(n((lnn)^2)) 1/(n((lnn)^2))
  • (2/9)^n (2/9)^n
  • n^2/2^n n^2/2^n

  • Identical expressions

  • one /(n((lnn)^ two))
  • 1 divide by (n((lnn) squared ))
  • one divide by (n((lnn) to the power of two))
  • 1/(n((lnn)2))
  • 1/nlnn2
  • 1/(n((lnn)²))
  • 1/(n((lnn) to the power of 2))
  • 1/nlnn^2
  • 1 divide by (n((lnn)^2))
Sum of series/ 1/(n((lnn)^2))

Sum of series 1/(n((lnn)^2))


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The solution

You have entered [src] 
  oo           
____           
\   `          
 \        1    
  \   ---------
  /        2   
 /    n*log (n)
/___,          
n = 2
$$\sum_{n=2}^{\infty} \frac{1}{n \log{\left(n \right)}^{2}}$$ 
Sum(1/(n*log(n)^2), (n, 2, 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((lnn)^2))

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