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  • Sum of series:
  • 1/((3n-2)(3n+1)) 1/((3n-2)(3n+1))
  • 3/(n(n+2)) 3/(n(n+2))
  • (n+1)/5^n (n+1)/5^n
  • 6/9n^2+12n-5 6/9n^2+12n-5

  • Identical expressions

  • arcsin(n+ one /n^ two)
  • arc sinus of (n plus 1 divide by n squared )
  • arc sinus of (n plus one divide by n to the power of two)
  • arcsin(n+1/n2)
  • arcsinn+1/n2
  • arcsin(n+1/n²)
  • arcsin(n+1/n to the power of 2)
  • arcsinn+1/n^2
  • arcsin(n+1 divide by n^2)

  • Similar expressions

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

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


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The solution

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

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