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Sum of series log5(n-i)



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The solution

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  oo            
 ___            
 \  `           
  \   log(n - i)
   )  ----------
  /     log(5)  
 /__,           
i = 0           
i=0log(i+n)log(5)\sum_{i=0}^{\infty} \frac{\log{\left(- i + n \right)}}{\log{\left(5 \right)}}
Sum(log(n - i)/log(5), (i, 0, oo))
The radius of convergence of the power series
Given number:
log(i+n)log(5)\frac{\log{\left(- i + n \right)}}{\log{\left(5 \right)}}
It is a series of species
ai(cxx0)dia_{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:
Rd=x0+limiaiai+1cR^{d} = \frac{x_{0} + \lim_{i \to \infty} \left|{\frac{a_{i}}{a_{i + 1}}}\right|}{c}
In this case
ai=log(i+n)log(5)a_{i} = \frac{\log{\left(- i + n \right)}}{\log{\left(5 \right)}}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=limilog((in))log((in+1))1 = \lim_{i \to \infty} \left|{\frac{\log{\left(- (i - n) \right)}}{\log{\left(- (i - n + 1) \right)}}}\right|
Let's take the limit
we find
True

False

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