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  • Sum of series:
  • x^n/n
  • 1/n! 1/n!
  • (9835^(2k/9835))/(9835+9835^(2k/9835)) (9835^(2k/9835))/(9835+9835^(2k/9835))
  • nx^n!

  • Identical expressions

  • (two *i+ one)/n^ two
  • (2 multiply by i plus 1) divide by n squared
  • (two multiply by i plus one) divide by n to the power of two
  • (2*i+1)/n2
  • 2*i+1/n2
  • (2*i+1)/n²
  • (2*i+1)/n to the power of 2
  • (2i+1)/n^2
  • (2i+1)/n2
  • 2i+1/n2
  • 2i+1/n^2
  • (2*i+1) divide by n^2

  • Similar expressions

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

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


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The solution

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

False
The answer [src] 
  2       2
pi    I*pi 
--- + -----
 6      3
$$\frac{\pi^{2}}{6} + \frac{i \pi^{2}}{3}$$ 
pi^2/6 + i*pi^2/3
Numerical answer [src] 
1.64493406684822643647241516665 + 3.28986813369645287294483033329*i
1.64493406684822643647241516665 + 3.28986813369645287294483033329*i

    Examples of finding the sum of a series

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