Evaluating $\sum_{k=0}^{\infty}\frac{1-2^k}{3^k}$

Question

I am trying to find the sum of following infinite series

$$\sum_{k=0}^{\infty}\frac{1-2^k}{3^k}$$

I tried starting the problem with rewriting it as $\dfrac13+\left(\frac{-2}{3}\right)^k$, am i on the right track at all?

Answer 1

Good Start but, It should be written as $$\sum_{k=0}^\infty\frac{1-2^k}{3^k}=\sum_{k=0}^\infty\left(\frac{1}{3}\right)^k-\sum_{k=0}^\infty\left(\frac{2}{3}\right)^k $$

Now both of these series are GP I hope you can take it from here!

$$\sum_{k=0}^\infty\left(\frac{1}{3}\right)^k=\frac{1}{1-\frac{1}{3}}=\frac32$$$$\sum_{k=0}^\infty\left(\frac{2}{3}\right)^k=\frac{1}{1-\frac23}=3$$

Answer 2

It should be rewritten as $(\frac{1}{3})^k-(\frac{2}{3})^k$