Question: After the great help I received (on here), I had an easy time working with a similar progression, but this one is giving me trouble. Can anyone sum the following:

$$\begin{align}
&0.3^{-1} \\
&0.3^{-1}+0.3^{0} \\
&0.3^{-1}+0.3^{0}+0.3^{1} \\
&0.3^{-1}+0.3^{0}+0.3^{1}+0.3^{2} \\
&\cdots
\end{align}$$ and so on.

I feel very thankful for all the help I've gotten on things thus far; thank you guys, your skills are epic and regarded with great respect, by me!

Answer: 
$\frac{\ln|3x|}{3}$ is *an* antiderivative of $\frac{1}{3x}$. But notice by properties of logarithms that
$$\frac{\ln|3x|}{3} = \frac{\ln(3|x|)}{3} = \frac{\ln 3 + \ln|x|}{3} = \frac{\ln|x|}{3} + \frac{\ln 3}{3}.$$

Also recall that you can add or subtract a constant from any antiderivative to obtain another antiderivative. So in particular, another antiderivative for $\frac{\ln|3x|}{3}$ is
$$\frac{\ln|x|}{3}+\frac{\ln 3}{3} - \frac{\ln 3}{3} = \frac{\ln|x|}{3}.$$
$$\frac{\ln(3x)}3=\frac{\ln x+\ln 3}3=\frac{\ln x}3+\frac{\ln 3}3$$
This function and
$$\frac{\ln x}3$$
have the same derivative.