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1/((4k^2)-1)

  • How to use it?

  • Sum of series:
  • 1/((4k^2)-1) 1/((4k^2)-1)
  • n*0,5^n n*0,5^n
  • (2/9)^n (2/9)^n
  • sin(n*pi/2)/n sin(n*pi/2)/n

  • Identical expressions

  • one /((4k^ two)- one)
  • 1 divide by ((4k squared ) minus 1)
  • one divide by ((4k to the power of two) minus one)
  • 1/((4k2)-1)
  • 1/4k2-1
  • 1/((4k²)-1)
  • 1/((4k to the power of 2)-1)
  • 1/4k^2-1
  • 1 divide by ((4k^2)-1)

  • Similar expressions

  • 1/((4k^2)+1)
Sum of series/ 1/((4k^2)-1)

Sum of series 1/((4k^2)-1)


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The solution

You have entered [src] 
  oo          
____          
\   `         
 \       1    
  \   --------
  /      2    
 /    4*k  - 1
/___,         
k = 1
$$\sum_{k=1}^{\infty} \frac{1}{4 k^{2} - 1}$$ 
Sum(1/(4*k^2 - 1), (k, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{4 k^{2} - 1}$$
It is a series of species
$$a_{k} \left(c x - x_{0}\right)^{d k}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{k \to \infty} \left|{\frac{a_{k}}{a_{k + 1}}}\right|}{c}$$
In this case
$$a_{k} = \frac{1}{4 k^{2} - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{k \to \infty}\left(\left(4 \left(k + 1\right)^{2} - 1\right) \left|{\frac{1}{4 k^{2} - 1}}\right|\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src] 
1/2
$$\frac{1}{2}$$ 
1/2
Numerical answer [src] 
0.500000000000000000000000000000
0.500000000000000000000000000000
The graph
Sum of series 1/((4k^2)-1)

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