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

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  • Sum of series:
  • 7x²
  • 10*60*0,18 10*60*0,18
  • tg^2×(1/n) tg^2×(1/n)
  • nln(n)/e^n nln(n)/e^n

  • Identical expressions

  • twelve /(n^ two - four *n+ three)
  • 12 divide by (n squared minus 4 multiply by n plus 3)
  • twelve divide by (n to the power of two minus four multiply by n plus 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 = 8
$$\sum_{n=8}^{\infty} \frac{12}{\left(n^{2} - 4 n\right) + 3}$$ 
Sum(12/(n^2 - 4*n + 3), (n, 8, 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
$$1 = \lim_{n \to \infty}\left(12 \left|{\frac{\frac{n}{3} - \frac{\left(n + 1\right)^{2}}{12} + \frac{1}{12}}{n^{2} - 4 n + 3}}\right|\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src] 
11/5
$$\frac{11}{5}$$ 
11/5
Numerical answer [src] 
2.20000000000000000000000000000
2.20000000000000000000000000000
The graph
Sum of series 12/(n^2-4*n+3)

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