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24/(9n^2-12n-5)
Sum of series/ 24/(9n^2-12n-5)

Sum of series 24/(9n^2-12n-5)


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The solution

You have entered [src] 
  oo                 
____                 
\   `                
 \           24      
  \   ---------------
  /      2           
 /    9*n  - 12*n - 5
/___,                
n = 1
$$\sum_{n=1}^{\infty} \frac{24}{\left(9 n^{2} - 12 n\right) - 5}$$ 
Sum(24/(9*n^2 - 12*n - 5), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{24}{\left(9 n^{2} - 12 n\right) - 5}$$
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
$$a_{n} = \frac{24}{9 n^{2} - 12 n - 5}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(24 \left|{\frac{\frac{n}{2} - \frac{3 \left(n + 1\right)^{2}}{8} + \frac{17}{24}}{- 9 n^{2} + 12 n + 5}}\right|\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src] 
3*Gamma(7/3)
------------
2*Gamma(4/3)
$$\frac{3 \Gamma\left(\frac{7}{3}\right)}{2 \Gamma\left(\frac{4}{3}\right)}$$ 
3*gamma(7/3)/(2*gamma(4/3))
Numerical answer [src] 
2.00000000000000000000000000000
2.00000000000000000000000000000
The graph
Sum of series 24/(9n^2-12n-5)

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