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6/9n^2+12n-5

  • How to use it?

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

  • Identical expressions

  • six /9n^ two +12n- five
  • 6 divide by 9n squared plus 12n minus 5
  • six divide by 9n to the power of two plus 12n minus five
  • 6/9n2+12n-5
  • 6/9n²+12n-5
  • 6/9n to the power of 2+12n-5
  • 6 divide by 9n^2+12n-5

  • Similar expressions

  • 6/9n^2-12n-5
  • 6/9n^2+12n+5
Sum of series/ 6/9n^2+12n-5

Sum of series 6/9n^2+12n-5


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The solution

You have entered [src] 
  oo                   
____                   
\   `                  
 \    /   2           \
  \   |2*n            |
  /   |---- + 12*n - 5|
 /    \ 3             /
/___,                  
n = 1
$$\sum_{n=1}^{\infty} \left(\left(\frac{2 n^{2}}{3} + 12 n\right) - 5\right)$$ 
Sum(2*n^2/3 + 12*n - 5, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\left(\frac{2 n^{2}}{3} + 12 n\right) - 5$$
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
$$a_{n} = \frac{2 n^{2}}{3} + 12 n - 5$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left|{\frac{2 n^{2}}{3} + 12 n - 5}\right|}{12 n + \frac{2 \left(n + 1\right)^{2}}{3} + 7}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
Numerical answer
The series diverges
The graph
Sum of series 6/9n^2+12n-5

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