sequences and series - Limit $sum_{k=0}^∞ left( sum_{j=0}^k binom{k}{j} left(-frac{1}{3}right)^j right)$

I have to find the limit of the following series:

$$\sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j} \left(-\frac{1}{3}\right)^j \right)$$

I don't even know how to approach this... Any help would be very appreciated

Answer

Using the binomial formula and the geometric series formula:

$$\sum_{k=0}^{\infty}\left(\sum_{j=0}^{k}{k\choose j}\left(-\frac13\right)^j\right)=\sum_{k=0}^{\infty}\left(1-\frac13\right)^k=\lim_{k\to\infty}\frac{(2/3)^{k+1}-1}{(2/3)-1}=\frac1{1-(2/3)}=3$$