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(8/((2*n+1)*(2*n+1)*3.1416*3.1416))*exp(-(2*n+1)*0.197*3.1416*3.1416*0.25)

Sum of series (8/((2*n+1)*(2*n+1)*3.1416*3.1416))*exp(-(2*n+1)*0.197*3.1416*3.1416*0.25)



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The solution

You have entered [src]
  oo                                                                  
_____                                                                 
\    `                                                                
 \                                        (-2*n - 1)*197              
  \                                       --------------*3.1416*3.1416
   \                                           1000                   
    )                                     ----------------------------
   /                   8                               4              
  /    ---------------------------------*e                            
 /     (2*n + 1)*(2*n + 1)*3.1416*3.1416                              
/____,                                                                
n = 0                                                                 
$$\sum_{n=0}^{\infty} \frac{8}{3.1416 \cdot 3.1416 \left(2 n + 1\right) \left(2 n + 1\right)} e^{\frac{3.1416 \cdot 3.1416 \frac{197 \left(- 2 n - 1\right)}{1000}}{4}}$$
Sum((8/(((((2*n + 1)*(2*n + 1))*3.1416)*3.1416)))*exp(((((-2*n - 1)*197/1000)*3.1416)*3.1416)/4), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\frac{8}{3.1416 \cdot 3.1416 \left(2 n + 1\right) \left(2 n + 1\right)} e^{\frac{3.1416 \cdot 3.1416 \frac{197 \left(- 2 n - 1\right)}{1000}}{4}}$$
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
$$a_{n} = \frac{0.498524174155581}{\left(2 n + 1\right)^{2}}$$
and
$$x_{0} = - e$$
,
$$d = -0.97216058016$$
,
$$c = 0$$
then
$$R^{-0.97216058016} = \tilde{\infty} \left(- e + \lim_{n \to \infty}\left(\frac{1 \left(2 n + 3\right)^{2}}{\left(2 n + 1\right)^{2}}\right)\right)$$
Let's take the limit
we find
False

$$R = 0$$
The rate of convergence of the power series
The answer [src]
0.405282839111986*polylog(2, 0.615032424328521) - 0.405282839111986*polylog(2, -0.615032424328521)
$$- 0.405282839111986 \operatorname{Li}_{2}\left(-0.615032424328521\right) + 0.405282839111986 \operatorname{Li}_{2}\left(0.615032424328521\right)$$
0.405282839111986*polylog(2, 0.615032424328521) - 0.405282839111986*polylog(2, -0.615032424328521)
Numerical answer [src]
0.523050803343365178221802125336
0.523050803343365178221802125336
The graph
Sum of series (8/((2*n+1)*(2*n+1)*3.1416*3.1416))*exp(-(2*n+1)*0.197*3.1416*3.1416*0.25)

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