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: $$S_n=\dfrac{a_1\cdot (1-r^n)}{1-r}$$ 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 $r=\dfrac{1}{3} < 1.$