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(2n+1)/(n(n+2)(n+4))

Sum of series (2n+1)/(n(n+2)(n+4))



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The solution

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  oo                   
 ___                   
 \  `                  
  \        2*n + 1     
   )  -----------------
  /   n*(n + 2)*(n + 4)
 /__,                  
n = 1                  
n=12n+1n(n+2)(n+4)\sum_{n=1}^{\infty} \frac{2 n + 1}{n \left(n + 2\right) \left(n + 4\right)}
Sum((2*n + 1)/(((n*(n + 2))*(n + 4))), (n, 1, oo))
The radius of convergence of the power series
Given number:
2n+1n(n+2)(n+4)\frac{2 n + 1}{n \left(n + 2\right) \left(n + 4\right)}
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=2n+1n(n+2)(n+4)a_{n} = \frac{2 n + 1}{n \left(n + 2\right) \left(n + 4\right)}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=limn((n+1)(n+3)(n+5)(2n+1)n(n+2)(n+4)(2n+3))1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 3\right) \left(n + 5\right) \left(2 n + 1\right)}{n \left(n + 2\right) \left(n + 4\right) \left(2 n + 3\right)}\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.01.0
The answer [src]
               0               0   
7      -1 + 4*e        -7 + 3*e    
-- + ------------- + --------------
12     /        0\      /        0\
     9*\-8 + 8*e /   12*\-8 + 8*e /
7+3e012(8+8e0)+712+1+4e09(8+8e0)\frac{-7 + 3 e^{0}}{12 \left(-8 + 8 e^{0}\right)} + \frac{7}{12} + \frac{-1 + 4 e^{0}}{9 \left(-8 + 8 e^{0}\right)}
7/12 + (-1 + 4*exp_polar(0))/(9*(-8 + 8*exp_polar(0))) + (-7 + 3*exp_polar(0))/(12*(-8 + 8*exp_polar(0)))
Numerical answer [src]
0.697916666666666666666666666667
0.697916666666666666666666666667
The graph
Sum of series (2n+1)/(n(n+2)(n+4))

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