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1/(4n^2-9)

  • How to use it?

  • Sum of series:
  • (2*n-1)/2^n (2*n-1)/2^n
  • (x-1)^n/5^n
  • 1/(4n^2-9) 1/(4n^2-9)
  • sin(n*x)/n^3

  • Identical expressions

  • one /(4n^ two - nine)
  • 1 divide by (4n squared minus 9)
  • one divide by (4n to the power of two minus nine)
  • 1/(4n2-9)
  • 1/4n2-9
  • 1/(4n²-9)
  • 1/(4n to the power of 2-9)
  • 1/4n^2-9
  • 1 divide by (4n^2-9)

  • Similar expressions

  • 1/(4n^2+9)
Sum of series/ 1/(4n^2-9)

Sum of series 1/(4n^2-9)


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The solution

You have entered [src] 
  oo          
____          
\   `         
 \       1    
  \   --------
  /      2    
 /    4*n  - 9
/___,         
n = 2
$$\sum_{n=2}^{\infty} \frac{1}{4 n^{2} - 9}$$ 
Sum(1/(4*n^2 - 9), (n, 2, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{4 n^{2} - 9}$$
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
$$a_{n} = \frac{1}{4 n^{2} - 9}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{4 \left(n + 1\right)^{2} - 9}{4 n^{2} - 9}}\right|$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src] 
           0                  0  
   -1 + 5*e          -14 + 8*e   
--------------- + ---------------
  /          0\     /          0\
4*\-15 + 15*e /   6*\-15 + 15*e /
$$\frac{-14 + 8 e^{0}}{6 \left(-15 + 15 e^{0}\right)} + \frac{-1 + 5 e^{0}}{4 \left(-15 + 15 e^{0}\right)}$$ 
(-1 + 5*exp_polar(0))/(4*(-15 + 15*exp_polar(0))) + (-14 + 8*exp_polar(0))/(6*(-15 + 15*exp_polar(0)))
Numerical answer [src] 
0.255555555555555555555555555556
0.255555555555555555555555555556
The graph
Sum of series 1/(4n^2-9)

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