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9/(9n^2-21n-8)

  • How to use it?

  • Sum of series:
  • (3/4)^n (3/4)^n
  • 1/((2n-1)*(2n+1)) 1/((2n-1)*(2n+1))
  • 1/(n*(n+3)) 1/(n*(n+3))
  • (5/6)^n (5/6)^n

  • Identical expressions

  • nine /(9n^ two -21n- eight)
  • 9 divide by (9n squared minus 21n minus 8)
  • nine divide by (9n to the power of two minus 21n minus eight)
  • 9/(9n2-21n-8)
  • 9/9n2-21n-8
  • 9/(9n²-21n-8)
  • 9/(9n to the power of 2-21n-8)
  • 9/9n^2-21n-8
  • 9 divide by (9n^2-21n-8)

  • Similar expressions

  • 9/(9n^2-21n+8)
  • 9/(9n^2+21n-8)
Sum of series/ 9/(9n^2-21n-8)

Sum of series 9/(9n^2-21n-8)


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The solution

You have entered [src] 
  oo                 
____                 
\   `                
 \           9       
  \   ---------------
  /      2           
 /    9*n  - 21*n - 8
/___,                
n = 0
$$\sum_{n=0}^{\infty} \frac{9}{\left(9 n^{2} - 21 n\right) - 8}$$ 
Sum(9/(9*n^2 - 21*n - 8), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\frac{9}{\left(9 n^{2} - 21 n\right) - 8}$$
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
$$a_{n} = \frac{9}{9 n^{2} - 21 n - 8}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(9 \left|{\frac{\frac{7 n}{3} - \left(n + 1\right)^{2} + \frac{29}{9}}{- 9 n^{2} + 21 n + 8}}\right|\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src] 
    /         0\                /        0\           
  9*\-32 + 8*e /*Gamma(4/3)   3*\27 + 9*e /*Gamma(4/3)
- ------------------------- - ------------------------
     /        0\                /        0\           
  16*\-5 + 5*e /*Gamma(1/3)   8*\-5 + 5*e /*Gamma(1/3)
$$- \frac{3 \left(9 e^{0} + 27\right) \Gamma\left(\frac{4}{3}\right)}{8 \left(-5 + 5 e^{0}\right) \Gamma\left(\frac{1}{3}\right)} - \frac{9 \left(-32 + 8 e^{0}\right) \Gamma\left(\frac{4}{3}\right)}{16 \left(-5 + 5 e^{0}\right) \Gamma\left(\frac{1}{3}\right)}$$ 
-9*(-32 + 8*exp_polar(0))*gamma(4/3)/(16*(-5 + 5*exp_polar(0))*gamma(1/3)) - 3*(27 + 9*exp_polar(0))*gamma(4/3)/(8*(-5 + 5*exp_polar(0))*gamma(1/3))
Numerical answer [src] 
-0.825000000000000000000000000000
-0.825000000000000000000000000000
The graph
Sum of series 9/(9n^2-21n-8)

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