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(2^(n+2))/(7^n*9^(n-1))

  • How to use it?

  • Sum of series:
  • (2^n+5^n)/10^n (2^n+5^n)/10^n
  • (2^(n+2))/(7^n*9^(n-1)) (2^(n+2))/(7^n*9^(n-1))
  • n/1+n^2 n/1+n^2
  • (x+4)^n/n^2

  • Identical expressions

  • (two ^(n+ two))/(seven ^n* nine ^(n- one))
  • (2 to the power of (n plus 2)) divide by (7 to the power of n multiply by 9 to the power of (n minus 1))
  • (two to the power of (n plus two)) divide by (seven to the power of n multiply by nine to the power of (n minus one))
  • (2(n+2))/(7n*9(n-1))
  • 2n+2/7n*9n-1
  • (2^(n+2))/(7^n9^(n-1))
  • (2(n+2))/(7n9(n-1))
  • 2n+2/7n9n-1
  • 2^n+2/7^n9^n-1
  • (2^(n+2)) divide by (7^n*9^(n-1))

  • Similar expressions

  • (2^(n+2))/(7^n*9^(n+1))
  • (2^(n-2))/(7^n*9^(n-1))
Sum of series/ (2^(n+2))/(7^n*9^(n-1))

Sum of series (2^(n+2))/(7^n*9^(n-1))


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The solution

You have entered [src] 
  oo           
____           
\   `          
 \       n + 2 
  \     2      
   )  ---------
  /    n  n - 1
 /    7 *9     
/___,          
n = 1
$$\sum_{n=1}^{\infty} \frac{2^{n + 2}}{7^{n} 9^{n - 1}}$$ 
Sum(2^(n + 2)/((7^n*9^(n - 1))), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{2^{n + 2}}{7^{n} 9^{n - 1}}$$
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
$$a_{n} = 2^{n + 2} \cdot 9^{1 - n}$$
and
$$x_{0} = -7$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-7 + \lim_{n \to \infty}\left(2^{- n - 3} \cdot 2^{n + 2} \cdot 9^{n} 9^{1 - n}\right)\right)$$
Let's take the limit
we find
False

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
72
--
61
$$\frac{72}{61}$$ 
72/61
Numerical answer [src] 
1.18032786885245901639344262295
1.18032786885245901639344262295
The graph
Sum of series (2^(n+2))/(7^n*9^(n-1))

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