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(3^n-2^n)/6^n

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  • Sum of series:
  • (2^n+(-1)^n)/5^n (2^n+(-1)^n)/5^n
  • (n+1)^2/2^(n-1) (n+1)^2/2^(n-1)
  • 3/(n*(n+2)) 3/(n*(n+2))
  • 1/(4n^2-1) 1/(4n^2-1)

  • Identical expressions

  • (three ^n- two ^n)/ six ^n
  • (3 to the power of n minus 2 to the power of n) divide by 6 to the power of n
  • (three to the power of n minus two to the power of n) divide by six to the power of n
  • (3n-2n)/6n
  • 3n-2n/6n
  • 3^n-2^n/6^n
  • (3^n-2^n) divide by 6^n

  • Similar expressions

  • (3^n+2^n)/6^n
Sum of series/ (3^n-2^n)/6^n

Sum of series (3^n-2^n)/6^n


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \     n    n
  \   3  - 2 
   )  -------
  /       n  
 /       6   
/___,        
n = 1
$$\sum_{n=1}^{\infty} \frac{- 2^{n} + 3^{n}}{6^{n}}$$ 
Sum((3^n - 2^n)/6^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{- 2^{n} + 3^{n}}{6^{n}}$$
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
$$a_{n} = - 2^{n} + 3^{n}$$
and
$$x_{0} = -6$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-6 + \lim_{n \to \infty} \left|{\frac{2^{n} - 3^{n}}{2^{n + 1} - 3^{n + 1}}}\right|\right)$$
Let's take the limit
we find
False

False

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
1/2
$$\frac{1}{2}$$ 
1/2
Numerical answer [src] 
0.500000000000000000000000000000
0.500000000000000000000000000000
The graph
Sum of series (3^n-2^n)/6^n

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