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1/(2*n-1)^4

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  • Sum of series:
  • ln(1-1/k^2)
  • 1/(2*n-1)^4 1/(2*n-1)^4
  • n^(2n)/n! n^(2n)/n!
  • 4x

  • Identical expressions

  • one /(two *n- one)^ four
  • 1 divide by (2 multiply by n minus 1) to the power of 4
  • one divide by (two multiply by n minus one) to the power of four
  • 1/(2*n-1)4
  • 1/2*n-14
  • 1/(2*n-1)⁴
  • 1/(2n-1)^4
  • 1/(2n-1)4
  • 1/2n-14
  • 1/2n-1^4
  • 1 divide by (2*n-1)^4

  • Similar expressions

  • 1/(2*n+1)^4
Sum of series/ 1/(2*n-1)^4

Sum of series 1/(2*n-1)^4


∑
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The solution

You have entered [src] 
  oo            
____            
\   `           
 \        1     
  \   ----------
  /            4
 /    (2*n - 1) 
/___,           
n = 1
∑n=1∞1(2n−1)4\sum_{n=1}^{\infty} \frac{1}{\left(2 n - 1\right)^{4}} 
Sum(1/((2*n - 1)^4), (n, 1, oo))
The radius of convergence of the power series
Given number:
1(2n−1)4\frac{1}{\left(2 n - 1\right)^{4}}
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(2n−1)4a_{n} = \frac{1}{\left(2 n - 1\right)^{4}}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞((2n+1)4∣1(2n−1)4∣)1 = \lim_{n \to \infty}\left(\left(2 n + 1\right)^{4} \left|{\frac{1}{\left(2 n - 1\right)^{4}}}\right|\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.981.02
The answer [src] 
  4
pi 
---
 96
π496\frac{\pi^{4}}{96} 
pi^4/96
Numerical answer [src] 
1.01467803160419205454625346551
1.01467803160419205454625346551
The graph
Sum of series 1/(2*n-1)^4

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