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Sum of series sin((2n-1)-x)/(2n-1)^2



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The solution

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

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