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

  • Identical expressions

  • seven ^x/(five ^x- three)
  • 7 to the power of x divide by (5 to the power of x minus 3)
  • seven to the power of x divide by (five to the power of x minus three)
  • 7x/(5x-3)
  • 7x/5x-3
  • 7^x/5^x-3
  • 7^x divide by (5^x-3)

  • Similar expressions

  • 7^x/(5^x+3)
Sum of series/ 7^x/(5^x-3)

Sum of series 7^x/(5^x-3)


∑
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The solution

You have entered [src] 
  oo        
____        
\   `       
 \       x  
  \     7   
   )  ------
  /    x    
 /    5  - 3
/___,       
n = 1
$$\sum_{n=1}^{\infty} \frac{7^{x}}{5^{x} - 3}$$ 
Sum(7^x/(5^x - 3), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{7^{x}}{5^{x} - 3}$$
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
$$a_{n} = \frac{7^{x}}{5^{x} - 3}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find
True

False
The answer [src] 
     x 
 oo*7  
-------
      x
-3 + 5
$$\frac{\infty 7^{x}}{5^{x} - 3}$$ 
oo*7^x/(-3 + 5^x)

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