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  • Sum of series:
  • 1/n^3 1/n^3
  • ln(1-1/k^2)
  • (cos(nx)*3^n)/2^n
  • 1/(2*n-1)^4 1/(2*n-1)^4

  • 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
∑n=2∞log⁡(1−1k2)\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⁡(1−1k2)\log{\left(1 - \frac{1}{k^{2}} \right)}
It is a series of species
an(cx−x0)dna_{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:
Rd=x0+lim⁡n→∞∣anan+1∣cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=log⁡(1−1k2)a_{n} = \log{\left(1 - \frac{1}{k^{2}} \right)}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞11 = \lim_{n \to \infty} 1
Let's take the limit
we find
True

False
The answer [src] 
      /    1 \
oo*log|1 - --|
      |     2|
      \    k /
∞log⁡(1−1k2)\infty \log{\left(1 - \frac{1}{k^{2}} \right)} 
oo*log(1 - 1/k^2)

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