Infinite sums of series

If instead of adding only the $n$ first terms of a succession, we want to add them all, we will write: $S = \sum_{n \geq 1} a_{n}$

To indicate that we are adding all the terms from the first one. This sum $S$ is called series.

If the succession which series we are calculating is a geometric progression, we can extend the formula: $S_{n} = \frac{a_{1} \cdot (1 - r^{n})}{1 - r}$ tending $n$ to infinity, two situations are possible,

${\begin{cases} si r \leq 1 \Rightarrow r^{n} \to \infty \\ si r < 1 \Rightarrow r^{n} \to 0 \end{cases}$

Then we have two choices:

In a geometric progression of ratio $r \geq 1$ , the sums $S_{n}$ grow arbitrarily as the value of $n$ increases, and it is said that they tend to infinity, or that the series is divergent.
On the contrary, a geometric progression of ratio $r < 1$ the sums $S_{n}$ stabilize and they increasingly approach the quantity: $S = \frac{a_{1}}{1 - r}$ which is what we will call sum of the series. In this case we will say that the series is convergent.

Continuing with the previous examples,

Example

$\sum_{n \geq 1} a_{n} = \sum_{n \geq 1} 3 \cdot 2^{n - 1}$ it is divergent because it is a series of a geometric progression of ratio $r = 2 \geq 1$ , while the series $\sum_{n \geq 1} b_{n} = \sum_{n \geq 1} \frac{7}{3^{n - 1}} = \frac{b_{1}}{1 - r} = \frac{7}{1 - \frac{1}{3}} = \frac{21}{2}$ is convergent since it is the series of a geometric progression of ratio $r = \frac{1}{3} < 1.$