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

  • How to use it?

  • Sum of series:
  • ln(1-(1/n^2)) ln(1-(1/n^2))
  • yi
  • cosnx/(n/(1/3))
  • e^x^2

  • Identical expressions

  • ln(one -(one /n^ two))
  • ln(1 minus (1 divide by n squared ))
  • ln(one minus (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-(1 divide by n^2))

  • Similar expressions

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

False
The rate of convergence of the power series
Numerical answer [src] 
-0.693147180559945309417232121458
-0.693147180559945309417232121458
The graph
Sum of series ln(1-(1/n^2))

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