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1/((4n-3)*(4n+1))
Sum of series/ 1/((4n-3)*(4n+1))

Sum of series 1/((4n-3)*(4n+1))


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The solution

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  oo                     
 ___                     
 \  `                    
  \            1         
   )  -------------------
  /   (4*n - 3)*(4*n + 1)
 /__,                    
n = 1
∑n=1∞1(4n−3)(4n+1)\sum_{n=1}^{\infty} \frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)} 
Sum(1/((4*n - 3)*(4*n + 1)), (n, 1, oo))
The radius of convergence of the power series
Given number:
1(4n−3)(4n+1)\frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)}
It is a series of species
an(cx−x0)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+lim⁡n→∞∣anan+1∣cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=1(4n−3)(4n+1)a_{n} = \frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞((4n+5)∣14n−3∣)1 = \lim_{n \to \infty}\left(\left(4 n + 5\right) \left|{\frac{1}{4 n - 3}}\right|\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.150.25
The answer [src] 
 Gamma(9/4) 
------------
5*Gamma(5/4)
Γ(94)5Γ(54)\frac{\Gamma\left(\frac{9}{4}\right)}{5 \Gamma\left(\frac{5}{4}\right)} 
gamma(9/4)/(5*gamma(5/4))
Numerical answer [src] 
0.250000000000000000000000000000
0.250000000000000000000000000000
The graph
Sum of series 1/((4n-3)*(4n+1))

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