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  • Sum of series:
  • n^(2/3)*arctg(1/n^2) n^(2/3)*arctg(1/n^2)
  • 1/4^1 1/4^1
  • 2n 2n
  • exp(-0.01*n) exp(-0.01*n)

  • 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

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
∑n=1∞x22x(x−14)\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:
x22x(x−14)x 2^{2 x} \left(x - \frac{1}{4}\right)
It is a series of species
an(cx−x0)dna_{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:
Rd=x0+lim⁡n→∞∣anan+1∣cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=22xx(x−14)a_{n} = 2^{2 x} x \left(x - \frac{1}{4}\right)
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞11 = \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)
∞22xx(x−14)\infty 2^{2 x} x \left(x - \frac{1}{4}\right) 
oo*x*2^(2*x)*(-1/4 + x)

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