Calculate $\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k$?

Question

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?

Answer 1

\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}

Answer 2

First multiply by $1^{n-k}$ $$\sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k\\ \sum\limits^\infty_{n=0}\sum\limits^n_{k=0} \binom n k(-\frac 1 3)^k1^{n-k}\\ \sum\limits^\infty_{n=0}\sum\limits^n_{k=0} (1+\frac{-1}{3})^n\\ \sum\limits^\infty_{n=0} (\frac{+2}{3})^n=\\ \sum\limits^\infty_{n=0}((\frac{+2}{3})^0+(\frac{+2}{3})^1+(\frac{+2}{3})^3+...+(\frac{+2}{3})^n+...)=\\\lim_{n \to \infty}\frac{1(1-(\frac23)^{n+1})}{1-\frac23}\\ $$

Answer 3

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 $$\sum_{n=0}^\infty \sum_{k=0}^n \binom nk \left(-\frac 13\right)^k=\color{red}3$$