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  • Sum of series:
  • (1+sin(x))/n
  • (1-5/n)^n (1-5/n)^n
  • (1)/(n*(n+1)*(n+2)) (1)/(n*(n+1)*(n+2))
  • ((n+1)*x^n)/4^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



=

The solution

You have entered [src]
  oo         
____         
\   `        
 \    2*I + 1
  \   -------
  /       2  
 /       n   
/___,        
n = 1        
n=11+2in2\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:
1+2in2\frac{1 + 2 i}{n^{2}}
It is a series of species
an(cxx0)dna_{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:
Rd=x0+limnanan+1cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=1+2in2a_{n} = \frac{1 + 2 i}{n^{2}}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=limn((n+1)2n2)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  
π26+iπ23\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