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(2^(n+1))/(3^(n-2)*7^n)
  • How to use it?

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  • 1/(2*n-1)*(2*n+1) 1/(2*n-1)*(2*n+1)
  • Identical expressions

  • (two ^(n+ one))/(three ^(n- two)* seven ^n)
  • (2 to the power of (n plus 1)) divide by (3 to the power of (n minus 2) multiply by 7 to the power of n)
  • (two to the power of (n plus one)) divide by (three to the power of (n minus two) multiply by seven to the power of n)
  • (2(n+1))/(3(n-2)*7n)
  • 2n+1/3n-2*7n
  • (2^(n+1))/(3^(n-2)7^n)
  • (2(n+1))/(3(n-2)7n)
  • 2n+1/3n-27n
  • 2^n+1/3^n-27^n
  • (2^(n+1)) divide by (3^(n-2)*7^n)
  • Similar expressions

  • (2^(n+1))/(3^(n+2)*7^n)
  • (2^(n-1))/(3^(n-2)*7^n)

Sum of series (2^(n+1))/(3^(n-2)*7^n)



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The solution

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

$$R = 0$$
The rate of convergence of the power series
The answer [src]
36
--
19
$$\frac{36}{19}$$
36/19
Numerical answer [src]
1.89473684210526315789473684211
1.89473684210526315789473684211
The graph
Sum of series (2^(n+1))/(3^(n-2)*7^n)

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