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

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  • Similar expressions

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

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


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The solution

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

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

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