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ln(1+1/n^2)

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  • Sum of series:
  • ln(1+1/n^2) ln(1+1/n^2)
  • 1/(nlnn) 1/(nlnn)
  • sinx
  • (7n^2+5n-1)^3n+2/4n^2+2 (7n^2+5n-1)^3n+2/4n^2+2

  • Identical expressions

  • ln(one + one /n^ two)
  • ln(1 plus 1 divide by n squared )
  • ln(one plus one divide by n to the power of two)
  • ln(1+1/n2)
  • ln1+1/n2
  • ln(1+1/n²)
  • ln(1+1/n to the power of 2)
  • ln1+1/n^2
  • ln(1+1 divide by n^2)

  • Similar expressions

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

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


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The solution

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

False
The rate of convergence of the power series
The answer [src] 
  oo             
____             
\   `            
 \       /    1 \
  \   log|1 + --|
  /      |     2|
 /       \    n /
/___,            
n = 1
$$\sum_{n=1}^{\infty} \log{\left(1 + \frac{1}{n^{2}} \right)}$$ 
Sum(log(1 + n^(-2)), (n, 1, oo))
Numerical answer [src] 
1.30184639860371267777043366301
1.30184639860371267777043366301
The graph
Sum of series ln(1+1/n^2)

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