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Sum of series log(n/2^i)



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  oo         
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\   `        
 \       /n \
  \   log|--|
  /      | i|
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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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