Does this series $\sum_{i=0}^n \frac{4}{3^n}$ diverge or converge? I a newbie to series, and I have not done too much yet. I have an exercise where I have basically to say if some series are convergent or divergent. If convergent, determine (and prove) the sum of the series.
This is the first series:
$$\sum_{i=0}^n \frac{4}{3^i}$$
I have heard about the ratio test, so I decided to try to apply it in this case to see if the series is or not convergent.
The ratio test is basically defined like this:
$$L = \lim{\left|\frac{a_{n+1}}{a_n}\right|}$$
Where $a_n$ is in this case $\frac{4}{3^n}$ and $a_{n+1}$ is consequently $\frac{4}{3^{n+1}}$. Thus, we have:
$$L = \lim{\left|\frac{\frac{4}{3^{n+1}}}{\frac{4}{3^n}}\right|} = \lim{\left|\frac{4 \cdot 3^n}{3^{n+1} \cdot 4}\right|} = \lim{\left|\frac{3^n}{3^{n+1}}\right|} = \lim{\left|\frac{3^n}{3^n\cdot 3}\right|} = \lim{\left|\frac{1}{3}\right|} = \frac{1}{3} < 1$$
From the ratio test, we know if $L< 1$, then the series converges.
Now, I need to find the sum. I have followed a video which explain how to find the sum using the ratio, which sincerely I have not understand well what it is. Can you explain a general rule to find the ratio, and what exactly is it?
Ok. So, following the same process done during the video, I have:
$$\sum_{i=0}^n \frac{4}{3^n} = 4 \sum_{i=0}^n \frac{1}{3^n} = 4 \sum_{i=0}^n \left(\frac{1}{3}\right)^n$$
Apparently, our ratio $r$ is $\frac{1}{3}$.
Now, to find the sum of the series, he uses a formula:
$$\frac{\text{first term}}{1 - r}$$
Where first term is the first term of the series. Can you explain where this formula comes from?
Applying that formula, we obtain:
$$\frac{\left(\frac{1}{3}\right)^0}{1 - \frac{1}{3}} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$$
I checked the sum using a online calculator, and apparently it is not $\frac{3}{2}$, so I decided to manipulate, and I thought I left a $4$ outside the series, so I multiply that $4$ by $\frac{3}{2}$, which produces $\frac{12}{2} = 6$, which is exactly the sum I found using 2 online calculators.
Now, my 3rd question is: why what I did is correct (or not correct)?

Let $S_N=\sum_{i=0}^{N} r^i=1+r+r^2+\cdots r^N$.  Note that 
$$\begin{align}
rS_N&=\sum_{i=0}^{N} r^{i+1}\\
&=r+r^2+r^3+\cdots+r^N+r^{N+1}
\end{align}$$
So, if we form the difference $S_N-rS_N$ we find $S_N-rS_N=(1-r)S_N=1-r^{N+1}$ where upon solving for $S_N$ we find 
$$\begin{align}
S_N&=\frac{1-r^{N+1}}{1-r}\\
\end{align}$$
If $r<1$, then $r^{N+1} \to 0$ as $N \to \infty$ and we have
$$\sum_{i=0}^{\infty} r^{i}=\frac{1}{1-r}$$.
If $r=\frac{1}{3}$, then $\sum_{i=0}^{\infty} (\frac13)^{i}=\frac{1}{1-\frac13}=\frac32$.
If the series is $4\sum_{i=0}^{\infty} (\frac13)^{i}$, then we have $4\sum_{i=0}^{\infty} (\frac13)^{i}=4\frac{1}{1-\frac13}=4\frac32 = 6$
