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3^(n+3)/(4^(n-1)*5^n)

  • How to use it?

  • Sum of series:
  • (5^n-2^n)/10^n (5^n-2^n)/10^n
  • 1/(n*ln(n)) 1/(n*ln(n))
  • 3^(n+3)/(4^(n-1)*5^n) 3^(n+3)/(4^(n-1)*5^n)
  • 1/(2*n+5)*(2*n+7) 1/(2*n+5)*(2*n+7)

  • Identical expressions

  • three ^(n+ three)/(four ^(n- one)* five ^n)
  • 3 to the power of (n plus 3) divide by (4 to the power of (n minus 1) multiply by 5 to the power of n)
  • three to the power of (n plus three) divide by (four to the power of (n minus one) multiply by five to the power of n)
  • 3(n+3)/(4(n-1)*5n)
  • 3n+3/4n-1*5n
  • 3^(n+3)/(4^(n-1)5^n)
  • 3(n+3)/(4(n-1)5n)
  • 3n+3/4n-15n
  • 3^n+3/4^n-15^n
  • 3^(n+3) divide by (4^(n-1)*5^n)

  • Similar expressions

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

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


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The solution

You have entered [src] 
  oo           
____           
\   `          
 \       n + 3 
  \     3      
   )  ---------
  /    n - 1  n
 /    4     *5 
/___,          
n = 1
$$\sum_{n=1}^{\infty} \frac{3^{n + 3}}{4^{n - 1} \cdot 5^{n}}$$ 
Sum(3^(n + 3)/((4^(n - 1)*5^n)), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{3^{n + 3}}{4^{n - 1} \cdot 5^{n}}$$
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
$$a_{n} = 3^{n + 3} \cdot 4^{1 - n}$$
and
$$x_{0} = -5$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-5 + \lim_{n \to \infty}\left(3^{- n - 4} \cdot 3^{n + 3} \cdot 4^{n} 4^{1 - n}\right)\right)$$
Let's take the limit
we find
False

False

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
324
---
 17
$$\frac{324}{17}$$ 
324/17
Numerical answer [src] 
19.0588235294117647058823529412
19.0588235294117647058823529412
The graph
Sum of series 3^(n+3)/(4^(n-1)*5^n)

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