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arcsin(1/n^2)

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

  • Identical expressions

  • arcsin(one /n^ two)
  • arc sinus of (1 divide by n squared )
  • arc sinus of (one divide by n to the power of two)
  • arcsin(1/n2)
  • arcsin1/n2
  • arcsin(1/n²)
  • arcsin(1/n to the power of 2)
  • arcsin1/n^2
  • arcsin(1 divide by n^2)
Sum of series/ arcsin(1/n^2)

Sum of series arcsin(1/n^2)


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The solution

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

False
The rate of convergence of the power series
The answer [src] 
  oo          
____          
\   `         
 \        /1 \
  \   asin|--|
  /       | 2|
 /        \n /
/___,         
n = 1
$$\sum_{n=1}^{\infty} \operatorname{asin}{\left(\frac{1}{n^{2}} \right)}$$ 
Sum(asin(n^(-2)), (n, 1, oo))
Numerical answer [src] 
2.21869835298393806208753996496
2.21869835298393806208753996496
The graph
Sum of series arcsin(1/n^2)

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