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

  • How to use it?

  • Sum of series:
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  • 5^n 5^n
  • 2(1/(n^2+5n+6)) 2(1/(n^2+5n+6))
  • ((-1)^n)*((x^n)/(2n*n!))

  • Identical expressions

  • (- one)^n*ln(one + one /(n^ two))
  • ( minus 1) to the power of n multiply by ln(1 plus 1 divide by (n squared ))
  • ( minus one) to the power of n multiply by ln(one plus one divide by (n to the power of two))
  • (-1)n*ln(1+1/(n2))
  • -1n*ln1+1/n2
  • (-1)^n*ln(1+1/(n²))
  • (-1) to the power of n*ln(1+1/(n to the power of 2))
  • (-1)^nln(1+1/(n^2))
  • (-1)nln(1+1/(n2))
  • -1nln1+1/n2
  • -1^nln1+1/n^2
  • (-1)^n*ln(1+1 divide by (n^2))

  • Similar expressions

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

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


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The solution

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

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