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((2x^2+7x)*(4)^(3x+1))/((x^2+3x+1)*64^x)
Sum of series/ ((2x^2+7x)*(4)^(3x+1))/((x^2+3x+1)*64^x)

Sum of series ((2x^2+7x)*(4)^(3x+1))/((x^2+3x+1)*64^x)


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The solution

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  oo                       
____                       
\   `                      
 \    /   2      \  3*x + 1
  \   \2*x  + 7*x/*4       
   )  ---------------------
  /     / 2          \   x 
 /      \x  + 3*x + 1/*64  
/___,                      
x = 1
$$\sum_{x=1}^{\infty} \frac{4^{3 x + 1} \left(2 x^{2} + 7 x\right)}{64^{x} \left(\left(x^{2} + 3 x\right) + 1\right)}$$ 
Sum(((2*x^2 + 7*x)*4^(3*x + 1))/(((x^2 + 3*x + 1)*64^x)), (x, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{4^{3 x + 1} \left(2 x^{2} + 7 x\right)}{64^{x} \left(\left(x^{2} + 3 x\right) + 1\right)}$$
It is a series of species
$$a_{x} \left(c x - x_{0}\right)^{d x}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{x \to \infty} \left|{\frac{a_{x}}{a_{x + 1}}}\right|}{c}$$
In this case
$$a_{x} = \frac{4^{3 x + 1} \left(2 x^{2} + 7 x\right)}{x^{2} + 3 x + 1}$$
and
$$x_{0} = -64$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-64 + \lim_{x \to \infty}\left(\frac{4^{- 3 x - 4} \cdot 4^{3 x + 1} \left(2 x^{2} + 7 x\right) \left(3 x + \left(x + 1\right)^{2} + 4\right)}{\left(7 x + 2 \left(x + 1\right)^{2} + 7\right) \left(x^{2} + 3 x + 1\right)}\right)\right)$$
Let's take the limit
we find
False

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
  oo                            
____                            
\   `                           
 \     1 + 3*x   -x /   2      \
  \   4       *64  *\2*x  + 7*x/
   )  --------------------------
  /               2             
 /           1 + x  + 3*x       
/___,                           
x = 1
$$\sum_{x=1}^{\infty} \frac{4^{3 x + 1} \cdot 64^{- x} \left(2 x^{2} + 7 x\right)}{x^{2} + 3 x + 1}$$ 
Sum(4^(1 + 3*x)*64^(-x)*(2*x^2 + 7*x)/(1 + x^2 + 3*x), (x, 1, oo))
Numerical answer
The series diverges
The graph
Sum of series ((2x^2+7x)*(4)^(3x+1))/((x^2+3x+1)*64^x)

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