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sin((2n+1)/(n^2*(n+1)^0.5))

  • How to use it?

  • Sum of series:
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  • 1/(2*n-1)*(2*n+1) 1/(2*n-1)*(2*n+1)
  • cos(kx)
  • ((-1)^(n-1))*(5^n+2)*x^n

  • Identical expressions

  • sin((two n+ one)/(n^2*(n+ one)^ zero . five))
  • sinus of ((2n plus 1) divide by (n squared multiply by (n plus 1) to the power of 0.5))
  • sinus of ((two n plus one) divide by (n squared multiply by (n plus one) to the power of zero . five))
  • sin((2n+1)/(n2*(n+1)0.5))
  • sin2n+1/n2*n+10.5
  • sin((2n+1)/(n²*(n+1)^0.5))
  • sin((2n+1)/(n to the power of 2*(n+1) to the power of 0.5))
  • sin((2n+1)/(n^2(n+1)^0.5))
  • sin((2n+1)/(n2(n+1)0.5))
  • sin2n+1/n2n+10.5
  • sin2n+1/n^2n+1^0.5
  • sin((2n+1) divide by (n^2*(n+1)^0.5))

  • Similar expressions

  • sin((2n-1)/(n^2*(n+1)^0.5))
  • sin((2n+1)/(n^2*(n-1)^0.5))
Sum of series/ sin((2n+1)/(n^2*(n+1)^0.5))

Sum of series sin((2n+1)/(n^2*(n+1)^0.5))


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The solution

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

False
The rate of convergence of the power series
The answer [src] 
  oo                   
____                   
\   `                  
 \       /  1 + 2*n   \
  \   sin|------------|
  /      | 2   _______|
 /       \n *\/ 1 + n /
/___,                  
n = 1
$$\sum_{n=1}^{\infty} \sin{\left(\frac{2 n + 1}{n^{2} \sqrt{n + 1}} \right)}$$ 
Sum(sin((1 + 2*n)/(n^2*sqrt(1 + n))), (n, 1, oo))
The graph
Sum of series sin((2n+1)/(n^2*(n+1)^0.5))

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