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lnn/(n((ln^4)(n+1)))
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  • Sum of series:
  • (2/5)^n (2/5)^n
  • (2n+1)/(n^2(n+1)^2) (2n+1)/(n^2(n+1)^2)
  • (5^n+4^n)/6^n (5^n+4^n)/6^n
  • sinx
  • Identical expressions

  • lnn/(n((ln^ four)(n+ one)))
  • lnn divide by (n((ln to the power of 4)(n plus 1)))
  • lnn divide by (n((ln to the power of four)(n plus one)))
  • lnn/(n((ln4)(n+1)))
  • lnn/nln4n+1
  • lnn/(n((ln⁴)(n+1)))
  • lnn/nln^4n+1
  • lnn divide by (n((ln^4)(n+1)))
  • Similar expressions

  • lnn/(n((ln^4)(n-1)))

Sum of series lnn/(n((ln^4)(n+1)))



=

The solution

You have entered [src]
  oo                   
____                   
\   `                  
 \          log(n)     
  \   -----------------
  /        4           
 /    n*log (n)*(n + 1)
/___,                  
n = 1                  
$$\sum_{n=1}^{\infty} \frac{\log{\left(n \right)}}{n \left(n + 1\right) \log{\left(n \right)}^{4}}$$
Sum(log(n)/((n*(log(n)^4*(n + 1)))), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\log{\left(n \right)}}{n \left(n + 1\right) \log{\left(n \right)}^{4}}$$
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
$$a_{n} = \frac{1}{n \left(n + 1\right) \log{\left(n \right)}^{3}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 2\right) \log{\left(n + 1 \right)}^{3} \left|{\frac{1}{\log{\left(n \right)}^{3}}}\right|}{n}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src]
  oo                   
____                   
\   `                  
 \            1        
  \   -----------------
  /                3   
 /    n*(1 + n)*log (n)
/___,                  
n = 1                  
$$\sum_{n=1}^{\infty} \frac{1}{n \left(n + 1\right) \log{\left(n \right)}^{3}}$$
Sum(1/(n*(1 + n)*log(n)^3), (n, 1, oo))
The graph
Sum of series lnn/(n((ln^4)(n+1)))

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