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24/(9n^2-12n+5)

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  • Sum of series:
  • 6/9n^2+12n-5 6/9n^2+12n-5
  • 5^n 5^n
  • 1/(3n+1)(3n+4) 1/(3n+1)(3n+4)
  • (3-sin*n)/n-lnn (3-sin*n)/n-lnn

  • Identical expressions

  • twenty-four /(9n^ two -12n+ five)
  • 24 divide by (9n squared minus 12n plus 5)
  • twenty minus four divide by (9n to the power of two minus 12n plus five)
  • 24/(9n2-12n+5)
  • 24/9n2-12n+5
  • 24/(9n²-12n+5)
  • 24/(9n to the power of 2-12n+5)
  • 24/9n^2-12n+5
  • 24 divide by (9n^2-12n+5)

  • Similar expressions

  • 24/(9n^2+12n+5)
  • 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
$$1 = \lim_{n \to \infty}\left(\left(- 12 n + 9 \left(n + 1\right)^{2} - 7\right) \left|{\frac{1}{9 n^{2} - 12 n + 5}}\right|\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
Numerical answer [src] 
14.8194952335788599394851486518
14.8194952335788599394851486518
The graph
Sum of series 24/(9n^2-12n+5)

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