find the taylor sum of sigma from 0 to infinity of ((-1)^n)(2)^(n +3)/(3^(2n + 1)
The general term is...
$\displaystyle a_{n}=(-1)^{n}\cdot \frac{2^{n+3}}{3^{2n+1}}= \frac{8}{3}\cdot (-1)^{n}\cdot (\frac{2}{9})^{n}$
... so that is a geometric series and is...
$\displaystyle \sum_{n=0}^{\infty}a_{n} = \frac{8}{3}\cdot \frac{1}{1+\frac{2}{9}}=\frac{24}{11}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n 2^{n+3}}{3^{2n+1}} = \sum_{n=0}^{\infty}\frac{(-1)^n\cdot 2^3\cdot 2^n}{3\cdot 3^{2n}} = \frac{8}{3}\sum_{n=0}^{\infty}\frac{(-1)^n 2^n}{(3^2)^n}$ $\displaystyle = \frac{8}{3}\sum_{n=0}^{\infty}\left(\frac{-2}{9}\right)^n = \frac{8}{3}\cdot\frac{1}{1-(-2/9)} = \frac{8}{3}\cdot\frac{9}{11} = \frac{72}{33} = \boxed{\frac{24}{11}}$
EDIT: I was beat to the chase!
If this is what you mean then
$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)}\cdot \frac{2^{n+3}}{3^{2n+1}}$
Then we can fix it up by rewriting it as
$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)}\cdot \frac{(\sqrt{2})^{2n+6}}{3^{2n+1}}$
$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}(4\sqrt{2})}{(2n+1)}\cdot \left( \frac{\sqrt{2}}{3} \right)^{2n+1} $
So we can identify this with the function
$\displaystyle f(x)= (4\sqrt{2})\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)}\cdot \left( x \right)^{2n+1} $
So now we can take the derivative to get
$\displaystyle f'(x)= (4\sqrt{2})\sum_{n=0}^{\infty}(-1)^{n}\cdot \left( x \right)^{2n} $
This is a geometric series so
$\displaystyle f'(x)= (4\sqrt{2})\sum_{n=0}^{\infty}(-1)^{n}\cdot \left( x \right)^{2n} =4\sqrt{2} \cdot \frac{1}{1+x^2}$
So integrating we get
$\displaystyle f(x)=4\sqrt{2}\tan^{-1}(x)$
So now just evaluate this at $\displaystyle x=\frac{\sqrt{2}}{3}$