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  • Sum of series:
  • sinx
  • ln(1-1/k^2)
  • (2^n+3^n)/6^n (2^n+3^n)/6^n
  • lnn lnn

  • Identical expressions

  • ln(one - one /k^ two)
  • ln(1 minus 1 divide by k squared )
  • ln(one minus one divide by k to the power of two)
  • ln(1-1/k2)
  • ln1-1/k2
  • ln(1-1/k²)
  • ln(1-1/k to the power of 2)
  • ln1-1/k^2
  • ln(1-1 divide by k^2)

  • Similar expressions

  • ln(1+1/k^2)
Sum of series/ ln(1-1/k^2)

Sum of series ln(1-1/k^2)


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The solution

You have entered [src] 
  oo             
____             
\   `            
 \       /    1 \
  \   log|1 - --|
  /      |     2|
 /       \    k /
/___,            
n = 2
$$\sum_{n=2}^{\infty} \log{\left(1 - \frac{1}{k^{2}} \right)}$$ 
Sum(log(1 - 1/k^2), (n, 2, oo))
The radius of convergence of the power series
Given number:
$$\log{\left(1 - \frac{1}{k^{2}} \right)}$$
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
$$a_{n} = \log{\left(1 - \frac{1}{k^{2}} \right)}$$
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] 
      /    1 \
oo*log|1 - --|
      |     2|
      \    k /
$$\infty \log{\left(1 - \frac{1}{k^{2}} \right)}$$ 
oo*log(1 - 1/k^2)

    Examples of finding the sum of a series

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