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log((n^2+1)/n^2)

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  • Sum of series:
  • 1/(n(n+2)) 1/(n(n+2))
  • (2n-1)/2^n (2n-1)/2^n
  • sin(3n)/(7n)^(1/5) sin(3n)/(7n)^(1/5)
  • arctg1/(2n^2) arctg1/(2n^2)

  • Identical expressions

  • log((n^ two + one)/n^ two)
  • logarithm of ((n squared plus 1) divide by n squared )
  • logarithm of ((n to the power of two plus one) divide by n to the power of two)
  • log((n2+1)/n2)
  • logn2+1/n2
  • log((n²+1)/n²)
  • log((n to the power of 2+1)/n to the power of 2)
  • logn^2+1/n^2
  • log((n^2+1) divide by n^2)

  • Similar expressions

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

Sum of series log((n^2+1)/n^2)


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The solution

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

False
The rate of convergence of the power series
Numerical answer [src] 
1.30184639860371267777043366301
1.30184639860371267777043366301
The graph
Sum of series log((n^2+1)/n^2)

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