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1/(ln(n+3))^n

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  • Sum of series:
  • 6/9n^2+12n-5 6/9n^2+12n-5
  • 5^n 5^n
  • n^n/2^(n+1) n^n/2^(n+1)
  • 1/(3n+1)(3n+4) 1/(3n+1)(3n+4)

  • Identical expressions

  • one /(ln(n+ three))^n
  • 1 divide by (ln(n plus 3)) to the power of n
  • one divide by (ln(n plus three)) to the power of n
  • 1/(ln(n+3))n
  • 1/lnn+3n
  • 1/lnn+3^n
  • 1 divide by (ln(n+3))^n

  • Similar expressions

  • 1/(ln(n-3))^n
Sum of series/ 1/(ln(n+3))^n

Sum of series 1/(ln(n+3))^n


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The solution

You have entered [src] 
  oo             
____             
\   `            
 \         1     
  \   -----------
  /      n       
 /    log (n + 3)
/___,            
n = 1
$$\sum_{n=1}^{\infty} \frac{1}{\log{\left(n + 3 \right)}^{n}}$$ 
Sum(1/(log(n + 3)^n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{\log{\left(n + 3 \right)}^{n}}$$
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
$$a_{n} = \log{\left(n + 3 \right)}^{- n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\log{\left(n + 3 \right)}^{- n} \log{\left(n + 4 \right)}^{n + 1}\right)$$
Let's take the limit
we find
False

False
The rate of convergence of the power series
The answer [src] 
  oo              
 ___              
 \  `             
  \      -n       
  /   log  (3 + n)
 /__,             
n = 1
$$\sum_{n=1}^{\infty} \log{\left(n + 3 \right)}^{- n}$$ 
Sum(log(3 + n)^(-n), (n, 1, oo))
Numerical answer [src] 
1.38981887063478834882904616365
1.38981887063478834882904616365
The graph
Sum of series 1/(ln(n+3))^n

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