- #1
Dell
- 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 -
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 -