Calculate $\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k$? Is there a way to calculate $\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k$?
If I would be able to calculate it I could find the solution at another task. So hopefully it is possible. I feel like I am missing out onto something here. Is ther a way to simplify the second part?
 A: \begin{align}\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k\left(-\frac 1 3\right)^k&=\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k 1^{n-k}\left(-\frac 1 3\right)^k\\
&=\sum_{n=0}^{\infty}\left(1-\frac13\right)^n \\
&= \sum_{n=0}^{\infty}\left(\frac23\right)^n \\
&= \frac{1}{1-\frac23} \\
&=3\end{align}
A: First multiply  by  $1^{n-k}$ $$\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k\\
\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k1^{n-k}\\
\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} (1+\frac{-1}{3})^n\\
\sum\limits^\infty_{n=0} (\frac{+2}{3})^n=\\
\sum\limits^\infty_{n=0}((\frac{+2}{3})^0+(\frac{+2}{3})^1+(\frac{+2}{3})^3+...+(\frac{+2}{3})^n+...)=\\\lim_{n \to \infty}\frac{1(1-(\frac23)^{n+1})}{1-\frac23}\\
$$
A: Note that
$$\sum_{n=0}^\infty \color{blue}{\sum_{k=0}^n \binom nk (-r)^k}=\sum_{n=0}^\infty \color{blue}{(1-r)^k}
=\frac 1{1-(1-r)}=\frac 1r$$
Putting $r=\frac 13$ gives
$$\sum_{n=0}^\infty \sum_{k=0}^n \binom nk \left(-\frac 13\right)^k=\color{red}3$$
