How to Use Integration and Derivatives to Find the Sum of a Series

[Dell]

_∞
Σ (x^2n-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*x^2n-1, but i need the format of the integration- a division format- but the numberss from the derivative- 2n-1 -

 

[dx]

Hint: Change the limits to n = 0 to ∞ and (2n - 1) to (2n + 1).

 

[Dell]

tell me if this is right

_∞
Σ x^2n-1/(2n-1)
ⁿ⁼¹
=
_∞
Σ x^2(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

\intfx=\int1/(1-x²)=\intΣ 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

 

[dx]

Ya looks good.