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ln(sqrt(n))/4^n

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  • Sum of series:
  • sinn/n^2 sinn/n^2
  • ln5 ln5
  • ln(sqrt(n))/4^n ln(sqrt(n))/4^n
  • 1/(2k+1)^2

  • Identical expressions

  • ln(sqrt(n))/ four ^n
  • ln( square root of (n)) divide by 4 to the power of n
  • ln( square root of (n)) divide by four to the power of n
  • ln(√(n))/4^n
  • ln(sqrt(n))/4n
  • lnsqrtn/4n
  • lnsqrtn/4^n
  • ln(sqrt(n)) divide by 4^n
Sum of series/ ln(sqrt(n))/4^n

Sum of series ln(sqrt(n))/4^n


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The solution

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

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
  oo                
 ___                
 \  `               
  \    -n    /  ___\
  /   4  *log\\/ n /
 /__,               
n = 1
$$\sum_{n=1}^{\infty} 4^{- n} \log{\left(\sqrt{n} \right)}$$ 
Sum(4^(-n)*log(sqrt(n)), (n, 1, oo))
Numerical answer [src] 
0.0340368668699901685928816555060
0.0340368668699901685928816555060
The graph
Sum of series ln(sqrt(n))/4^n

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