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(3^n-4^n)/12^n

  • How to use it?

  • Sum of series:
  • 1/(n(n+3)) 1/(n(n+3))
  • (3^n-1)/n! (3^n-1)/n!
  • (3^n-4^n)/12^n (3^n-4^n)/12^n
  • 1/((n+14)(n+15)) 1/((n+14)(n+15))

  • Identical expressions

  • (three ^n- four ^n)/ twelve ^n
  • (3 to the power of n minus 4 to the power of n) divide by 12 to the power of n
  • (three to the power of n minus four to the power of n) divide by twelve to the power of n
  • (3n-4n)/12n
  • 3n-4n/12n
  • 3^n-4^n/12^n
  • (3^n-4^n) divide by 12^n

  • Similar expressions

  • (3^n+4^n)/12^n
Sum of series/ (3^n-4^n)/12^n

Sum of series (3^n-4^n)/12^n


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \     n    n
  \   3  - 4 
   )  -------
  /       n  
 /      12   
/___,        
n = 1
$$\sum_{n=1}^{\infty} \frac{3^{n} - 4^{n}}{12^{n}}$$ 
Sum((3^n - 4^n)/12^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{3^{n} - 4^{n}}{12^{n}}$$
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
$$a_{n} = 3^{n} - 4^{n}$$
and
$$x_{0} = -12$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-12 + \lim_{n \to \infty} \left|{\frac{3^{n} - 4^{n}}{3^{n + 1} - 4^{n + 1}}}\right|\right)$$
Let's take the limit
we find
False

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
-1/6
$$- \frac{1}{6}$$ 
-1/6
Numerical answer [src] 
-0.166666666666666666666666666667
-0.166666666666666666666666666667
The graph
Sum of series (3^n-4^n)/12^n

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