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Sum of series/ arctan(1/n^2+n+1)

Sum of series arctan(1/n^2+n+1)


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The solution

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

False
The answer [src] 
  oo                  
____                  
\   `                 
 \        /        1 \
  \   atan|1 + n + --|
  /       |         2|
 /        \        n /
/___,                 
n = 0
$$\sum_{n=0}^{\infty} \operatorname{atan}{\left(n + 1 + \frac{1}{n^{2}} \right)}$$ 
Sum(atan(1 + n + n^(-2)), (n, 0, oo))
Numerical answer [src] 
Sum(atan(1/(n^2) + n + 1), (n, 0, oo))
Sum(atan(1/(n^2) + n + 1), (n, 0, oo))

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