<s>[INST] 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)?
 [/INST] Here is a simple proof for the geometric convergence:
Suppose we have the finite sum:
$$1+r+r^2+r^3+r^4+\dots+r^k$$ with $r\neq 1$
Since this is a finite sum, it will converge to some unknown quantity: $S$
So:
$$\begin{array}{lrl} S & =& 1+r+r^2+r^3+r^4+\dots+r^k\\
r\cdot S & =& ~~~~~~~ r+r^2+r^3+r^4+r^5+\dots+r^{k+1}\\
S-r\cdot S& = & 1 - r^{k+1}\\
S & = & \frac{1-r^{k+1}}{1-r}\end{array}$$
If $r=1$ then clearly $S = k+1$
Using this information for infinite series, it follows that $\sum\limits_{i=0}^\infty r^i = \lim\limits_{k\to\infty}\sum\limits_{i=0}^k r^i = \lim\limits_{k\to\infty} \frac{1-r^{k+1}}{1-r}$
In the case that $|r|>1$ this clearly diverges, and when $|r|<1$ then the $r^{k+1}$ term is insignificant, so this will converge.  The case where $r=1$ followed the other form where $S = k+1$ and will clearly diverge, the case where $r=-1$ will also diverge since the sequence of partial sums will alternate between 1 and 0.
As for a general rule to find the ratio, in a geometric series, it will be of the form $\sum a \cdot r^n$.  Try to algebraically manipulate the summand to be in that form.  For example: $\sum 3\cdot \frac{5^n}{6^{n+1}} = \sum 3\cdot \frac{5^n}{6^n\cdot 6} = \sum \frac{3}{6}\cdot (\frac{5}{6})^n$, and your ratio will be $\frac{5}{6}$
</s>