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ln(1/(1/n^2))

  • How to use it?

  • Sum of series:
  • 1/x^n
  • sqrt(n^2+3n)-sqrt(n^2-n) sqrt(n^2+3n)-sqrt(n^2-n)
  • sqrt(n)/(n+3) sqrt(n)/(n+3)
  • ln(1/(1/n^2)) ln(1/(1/n^2))

  • Identical expressions

  • ln(one /(one /n^ two))
  • ln(1 divide by (1 divide by n squared ))
  • ln(one divide by (one divide by n to the power of two))
  • ln(1/(1/n2))
  • ln1/1/n2
  • ln(1/(1/n²))
  • ln(1/(1/n to the power of 2))
  • ln1/1/n^2
  • ln(1 divide by (1 divide by n^2))
Sum of series/ ln(1/(1/n^2))

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


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The solution

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

False
The rate of convergence of the power series
The answer [src] 
  oo         
 ___         
 \  `        
  \      / 2\
  /   log\n /
 /__,        
n = 2
$$\sum_{n=2}^{\infty} \log{\left(n^{2} \right)}$$ 
Sum(log(n^2), (n, 2, oo))
Numerical answer
The series diverges
The graph
Sum of series ln(1/(1/n^2))

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