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: $$$S_n=\dfrac{a_1\cdot (1-r^n)}{1-r}$$$ tending $$n$$ to infinity, two situations are possible,
$$$\left\{ \begin{array}{l} \mbox{si} \ r\leq 1 \Rightarrow r^n\rightarrow \infty \\\\ \mbox{si} \ r < 1 \Rightarrow r^n\rightarrow 0 \end{array} \right.$$$
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: $$$S=\dfrac{a_1}{1-r}$$$ 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,
$$\sum_{n \geq 1}a_n = \sum_{n \geq 1}3\cdot 2^{n-1}$$ it is divergent because it is a series of a geometric progression of ratio $$r=2 \geq 1$$, while the series $$\sum_{n \geq 1}b_n = \sum_{n \geq 1}\dfrac{7}{3^{n-1}}=\dfrac{b_1}{1-r}=\dfrac{7}{1-\dfrac{1}{3}}=\dfrac{21}{2}$$ is convergent since it is the series of a geometric progression of ratio $$r=\dfrac{1}{3} < 1.$$