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12/(4n^2+12n+5)

Sum of series 12/(4n^2+12n+5)



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The solution

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  oo                 
____                 
\   `                
 \           12      
  \   ---------------
  /      2           
 /    4*n  + 12*n + 5
/___,                
n = 1                
$$\sum_{n=1}^{\infty} \frac{12}{\left(4 n^{2} + 12 n\right) + 5}$$
Sum(12/(4*n^2 + 12*n + 5), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{12}{\left(4 n^{2} + 12 n\right) + 5}$$
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
$$a_{n} = \frac{12}{4 n^{2} + 12 n + 5}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{12 \left(n + \frac{\left(n + 1\right)^{2}}{3} + \frac{17}{12}\right)}{4 n^{2} + 12 n + 5}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src]
8/5
$$\frac{8}{5}$$
8/5
Numerical answer [src]
1.60000000000000000000000000000
1.60000000000000000000000000000
The graph
Sum of series 12/(4n^2+12n+5)

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