- #1

Dell

- 590

- 0

_{∞}

Σ (x

^{2n-1})/(2n-1)

^{n=1}

what i tried to do was take the taylor series:

_{∞}

Σ (x

^{n})=1/(1-x)

^{n=1}

so i can substitute x

^{2}for x

_{∞}

Σ (x

^{2n})=1/(1-x

^{2})

^{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 x

^{2n+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 -