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  • Sum of series:
  • 2^n/n^2 2^n/n^2
  • (n-1)/2^(n+1) (n-1)/2^(n+1)
  • tan(n*x)
  • arctg/n^2 arctg/n^2
  • Identical expressions

  • two ^(2x)(x- one / four)x
  • 2 to the power of (2x)(x minus 1 divide by 4)x
  • two to the power of (2x)(x minus one divide by four)x
  • 2(2x)(x-1/4)x
  • 22xx-1/4x
  • 2^2xx-1/4x
  • 2^(2x)(x-1 divide by 4)x
  • Similar expressions

  • 2^(2x)(x+1/4)x

Sum of series 2^(2x)(x-1/4)x



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The solution

You have entered [src]
  oo                  
 ___                  
 \  `                 
  \    2*x            
  /   2   *(x - 1/4)*x
 /__,                 
n = 1                 
$$\sum_{n=1}^{\infty} x 2^{2 x} \left(x - \frac{1}{4}\right)$$
Sum((2^(2*x)*(x - 1/4))*x, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$x 2^{2 x} \left(x - \frac{1}{4}\right)$$
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
$$a_{n} = 2^{2 x} x \left(x - \frac{1}{4}\right)$$
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]
      2*x           
oo*x*2   *(-1/4 + x)
$$\infty 2^{2 x} x \left(x - \frac{1}{4}\right)$$
oo*x*2^(2*x)*(-1/4 + x)

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