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ln^2(n)*ln(1+1/n)

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  • Sum of series:
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  • x^n/sqrt(n+1)
  • 12/(4n^2+12n+5) 12/(4n^2+12n+5)
  • 1/(sqrt(k^4+7k^2+20))

  • Identical expressions

  • ln^ two (n)*ln(one + one /n)
  • ln squared (n) multiply by ln(1 plus 1 divide by n)
  • ln to the power of two (n) multiply by ln(one plus one divide by n)
  • ln2(n)*ln(1+1/n)
  • ln2n*ln1+1/n
  • ln²(n)*ln(1+1/n)
  • ln to the power of 2(n)*ln(1+1/n)
  • ln^2(n)ln(1+1/n)
  • ln2(n)ln(1+1/n)
  • ln2nln1+1/n
  • ln^2nln1+1/n
  • ln^2(n)*ln(1+1 divide by n)

  • Similar expressions

  • ln^2(n)*ln(1-1/n)
Sum of series/ ln^2(n)*ln(1+1/n)

Sum of series ln^2(n)*ln(1+1/n)


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The solution

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

False
The rate of convergence of the power series
The graph
Sum of series ln^2(n)*ln(1+1/n)

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