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12/(n^2-4*n-3)

  • How to use it?

  • Sum of series:
  • seno^2x(2x)
  • arcsin1/sqrtn arcsin1/sqrtn
  • 1/(2n-1)*(2n+2) 1/(2n-1)*(2n+2)
  • 1/((4*n+2)*(4*n+1)) 1/((4*n+2)*(4*n+1))

  • Identical expressions

  • twelve /(n^ two - four *n- three)
  • 12 divide by (n squared minus 4 multiply by n minus 3)
  • twelve divide by (n to the power of two minus four multiply by n minus three)
  • 12/(n2-4*n-3)
  • 12/n2-4*n-3
  • 12/(n²-4*n-3)
  • 12/(n to the power of 2-4*n-3)
  • 12/(n^2-4n-3)
  • 12/(n2-4n-3)
  • 12/n2-4n-3
  • 12/n^2-4n-3
  • 12 divide by (n^2-4*n-3)

  • Similar expressions

  • 12/(n^2+4*n-3)
  • 12/(n^2-4*n+3)
Sum of series/ 12/(n^2-4*n-3)

Sum of series 12/(n^2-4*n-3)


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The solution

You have entered [src] 
  oo              
____              
\   `             
 \         12     
  \   ------------
  /    2          
 /    n  - 4*n - 3
/___,             
n = 4
$$\sum_{n=4}^{\infty} \frac{12}{\left(n^{2} - 4 n\right) - 3}$$ 
Sum(12/(n^2 - 4*n - 3), (n, 4, oo))
The radius of convergence of the power series
Given number:
$$\frac{12}{\left(n^{2} - 4 n\right) - 3}$$
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
$$a_{n} = \frac{12}{n^{2} - 4 n - 3}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(12 \left|{\frac{\frac{n}{3} - \frac{\left(n + 1\right)^{2}}{12} + \frac{7}{12}}{- n^{2} + 4 n + 3}}\right|\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
Numerical answer [src] 
6.36825693054860763306576988215
6.36825693054860763306576988215
The graph
Sum of series 12/(n^2-4*n-3)

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