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sqrt(n)*sin(pi/n^2)

  • How to use it?

  • Sum of series:
  • ln(1-(1/n^2)) ln(1-(1/n^2))
  • sqrt(n)*sin(pi/n^2) sqrt(n)*sin(pi/n^2)
  • e^x^2
  • cosnx/(n/(1/3))

  • Identical expressions

  • sqrt(n)*sin(pi/n^ two)
  • square root of (n) multiply by sinus of ( Pi divide by n squared )
  • square root of (n) multiply by sinus of ( Pi divide by n to the power of two)
  • √(n)*sin(pi/n^2)
  • sqrt(n)*sin(pi/n2)
  • sqrtn*sinpi/n2
  • sqrt(n)*sin(pi/n²)
  • sqrt(n)*sin(pi/n to the power of 2)
  • sqrt(n)sin(pi/n^2)
  • sqrt(n)sin(pi/n2)
  • sqrtnsinpi/n2
  • sqrtnsinpi/n^2
  • sqrt(n)*sin(pi divide by n^2)
Sum of series/ sqrt(n)*sin(pi/n^2)

Sum of series sqrt(n)*sin(pi/n^2)


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The solution

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

False
The rate of convergence of the power series
Numerical answer [src] 
4.93872735086760375010342509187
4.93872735086760375010342509187
The graph
Sum of series sqrt(n)*sin(pi/n^2)

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