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Find the sum of this series

  • Thread starter Dell
  • Start date
  • #1
590
0

Σ (x2n-1)/(2n-1)
n=1

what i tried to do was take the taylor series:


Σ (xn)=1/(1-x)
n=1

so i can substitute x2 for x


Σ (x2n)=1/(1-x2)
n=1


now i need to to use some form of integration/derivative to get to my series from the taylor series,
my problem is that integral will give me x2n+1/(2n+1) , and derivative will give me 2n*x2n-1, but i need the format of the integration- a division format- but the numberss from the derivative- 2n-1 -
 

Answers and Replies

  • #2
dx
Homework Helper
Gold Member
2,011
18
Hint: Change the limits to n = 0 to ∞ and (2n - 1) to (2n + 1).
 
  • #3
590
0
tell me if this is right


Σ x2n-1/(2n-1)
n=1
=

Σ x2(n+1)-1/(2(n+1)-1)
n=0
=

Σ x2n+1/(2n+1)
n=0

fx= 1/(1-x) = Σ xn
x==>x2
fx= 1/(1-x2) = Σ x2n

[tex]\int[/tex]fx=[tex]\int[/tex]1/(1-x2)=[tex]\int[/tex]Σ x2n

0.5*ln|(1+x)/(1-x)| = x2n+1/2n+1

and i can do all of this because i say n=1, and n*=0, therefore n* goes from 0-∞ and n goes from 1-∞, making n*=n+1 so i can use the same taylor series, but instead of n, i write n+1
 
  • #4
dx
Homework Helper
Gold Member
2,011
18
Ya looks good.
 

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