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sqrt(2/(n^3+4))

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  • Sum of series:
  • 1/n! 1/n!
  • ln(1-1/n^2) ln(1-1/n^2)
  • cos*π*n cos*π*n
  • cosn/n cosn/n

  • Identical expressions

  • sqrt(two /(n^ three + four))
  • square root of (2 divide by (n cubed plus 4))
  • square root of (two divide by (n to the power of three plus four))
  • √(2/(n^3+4))
  • sqrt(2/(n3+4))
  • sqrt2/n3+4
  • sqrt(2/(n³+4))
  • sqrt(2/(n to the power of 3+4))
  • sqrt2/n^3+4
  • sqrt(2 divide by (n^3+4))

  • Similar expressions

  • sqrt(2/(n^3-4))
Sum of series/ sqrt(2/(n^3+4))

Sum of series sqrt(2/(n^3+4))


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The solution

You have entered [src] 
  oo               
____               
\   `              
 \         ________
  \       /   2    
   )     /  ------ 
  /     /    3     
 /    \/    n  + 4 
/___,              
n = 1
$$\sum_{n=1}^{\infty} \sqrt{\frac{2}{n^{3} + 4}}$$ 
Sum(sqrt(2/(n^3 + 4)), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\sqrt{\frac{2}{n^{3} + 4}}$$
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
$$a_{n} = \sqrt{2} \sqrt{\frac{1}{n^{3} + 4}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt{\left(n + 1\right)^{3} + 4}}{\sqrt{n^{3} + 4}}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src] 
  oo              
_____             
\    `            
 \          ___   
  \       \/ 2    
   \   -----------
   /      ________
  /      /      3 
 /     \/  4 + n  
/____,            
n = 1
$$\sum_{n=1}^{\infty} \frac{\sqrt{2}}{\sqrt{n^{3} + 4}}$$ 
Sum(sqrt(2)/sqrt(4 + n^3), (n, 1, oo))
The graph
Sum of series sqrt(2/(n^3+4))

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