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  • Sum of series:
  • (4x)^(2n)
  • sinx
  • log(n/2^i)
  • 1/(r+1) 1/(r+1)

  • Identical expressions

  • log(n/ two ^i)
  • logarithm of (n divide by 2 to the power of i)
  • logarithm of (n divide by two to the power of i)
  • log(n/2i)
  • logn/2i
  • logn/2^i
  • log(n divide by 2^i)
Sum of series/ log(n/2^i)

Sum of series log(n/2^i)


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \       /n \
  \   log|--|
  /      | i|
 /       \2 /
/___,        
i = 0
$$\sum_{i=0}^{\infty} \log{\left(\frac{n}{2^{i}} \right)}$$ 
Sum(log(n/2^i), (i, 0, oo))
The radius of convergence of the power series
Given number:
$$\log{\left(\frac{n}{2^{i}} \right)}$$
It is a series of species
$$a_{i} \left(c x - x_{0}\right)^{d i}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{i \to \infty} \left|{\frac{a_{i}}{a_{i + 1}}}\right|}{c}$$
In this case
$$a_{i} = \log{\left(2^{- i} n \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{i \to \infty} \left|{\frac{\log{\left(2^{- i} n \right)}}{\log{\left(2^{- i - 1} n \right)}}}\right|$$
Let's take the limit
we find
True

False
The answer [src] 
  oo            
 ___            
 \  `           
  \      /   -i\
  /   log\n*2  /
 /__,           
i = 0
$$\sum_{i=0}^{\infty} \log{\left(2^{- i} n \right)}$$ 
Sum(log(n*2^(-i)), (i, 0, oo))

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