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sqrt(n)sin(1/(2n))^2
  • How to use it?

  • Sum of series:
  • 1/(2^n) 1/(2^n)
  • cos(n)*x/n^2
  • ((-1)^(n+1))/ln(n+1) ((-1)^(n+1))/ln(n+1)
  • sqrt(n)sin(1/(2n))^2 sqrt(n)sin(1/(2n))^2
  • Identical expressions

  • sqrt(n)sin(one /(two n))^2
  • square root of (n) sinus of (1 divide by (2n)) squared
  • square root of (n) sinus of (one divide by (two n)) squared
  • √(n)sin(1/(2n))^2
  • sqrt(n)sin(1/(2n))2
  • sqrtnsin1/2n2
  • sqrt(n)sin(1/(2n))²
  • sqrt(n)sin(1/(2n)) to the power of 2
  • sqrtnsin1/2n^2
  • sqrt(n)sin(1 divide by (2n))^2

Sum of series sqrt(n)sin(1/(2n))^2



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The solution

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

False
The rate of convergence of the power series
Numerical answer [src]
0.630319826453614403093174316242
0.630319826453614403093174316242
The graph
Sum of series sqrt(n)sin(1/(2n))^2

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