Summing $\sum_{k=1}^{\infty} \frac{(-2)^k + 1}{3^k}$

Question

I’m trying to calculate the sum of this by breaking it down into two geometric series:

$$\sum_{k=1}^{\infty} \frac{(-1)^k2^k + 1}{(3^k)} =\sum_{k=1}^\infty\left(-\frac{2}{3}\right)^{k} +\sum_{k=1}^\infty \left({\frac13}\right)^{k}$$

But I couldn’t proceed further and would like to get some help! thanks

Answer 1

My Solution:

\begin{align} &\sum_{k=0}^{\infty}\frac{(-1)^k2^k+1}{3^k}\\ =&\left[\sum_{k=0}^{\infty}(-\frac{2}{3})^k\right]+\left[\sum_{k=0}^{\infty}(\frac{1}{3})^k\right]\\ =&\frac{1}{1-(-\frac{2}{3})}+\frac{1}{1-\frac{1}{3}}\\ =&\frac{21}{10}. \end{align}

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Update: (starting from $k=1$)

\begin{align} &\sum_{k=1}^{\infty}\frac{(-1)^k2^k+1}{3^k}\\ =&\left[\sum_{k=1}^{\infty}(-\frac{2}{3})^k\right]+\left[\sum_{k=1}^{\infty}(\frac{1}{3})^k\right]\\ =&\frac{-\frac{2}{3}}{1-(-\frac{2}{3})}+\frac{\frac{1}{3}}{1-\frac{1}{3}}\\ =&\frac{1}{10}. \end{align}