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(-1)^n/n^2

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  • Sum of series:
  • (-1)^n/n^2 (-1)^n/n^2
  • lnn/n^2 lnn/n^2
  • sin(pi*sqrt(n^2+k^2))
  • ln(n/(n+1)) ln(n/(n+1))
  • Limit of the function:
  • (-1)^n/n^2 (-1)^n/n^2

  • Identical expressions

  • (- one)^n/n^ two
  • ( minus 1) to the power of n divide by n squared
  • ( minus one) to the power of n divide by n to the power of two
  • (-1)n/n2
  • -1n/n2
  • (-1)^n/n²
  • (-1) to the power of n/n to the power of 2
  • -1^n/n^2
  • (-1)^n divide by n^2

  • Similar expressions

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

Sum of series (-1)^n/n^2


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The solution

You have entered [src] 
  oo       
____       
\   `      
 \        n
  \   (-1) 
   )  -----
  /      2 
 /      n  
/___,      
n = 1
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{n}}{n^{2}}$$ 
Sum((-1)^n/n^2, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(-1\right)^{n}}{n^{2}}$$
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
$$a_{n} = \frac{1}{n^{2}}$$
and
$$x_{0} = 1$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(1 + \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)\right)$$
Let's take the limit
we find
False
The rate of convergence of the power series
The answer [src] 
   2 
-pi  
-----
  12
$$- \frac{\pi^{2}}{12}$$ 
-pi^2/12
Numerical answer [src] 
-0.822467033424113218236207583323
-0.822467033424113218236207583323
The graph
Sum of series (-1)^n/n^2

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