Example 4.3. Investigate the convergence of the series $\sum_{n=0}^{\infty} a^{n}$ (infinite geometric progression) $a \in R$.
Solution. Using the formula for the sum of the first terms of a geometric progression, we will calculate the partial sums of this series:

$$
\begin{gathered}
S_{n}=\frac{1-a^{n}}{1-a}=\frac{1}{1-a}-\frac{a^{n}}{1-a}(a \neq 1) \\
S_{n}=n(a=1)
\end{gathered}
$$

Thus, the question of the convergence of an infinite geometric progression reduces to the question of the existence of the limit

$$
\lim _{n \rightarrow \infty} \frac{a^{n}}{1-a}=\left\{\begin{array}{r}
0, \text { if }|a|<1 \\
\text { does not exist, if } a \leqslant-1
\end{array}\right.
$$

To summarize: an infinite geometric progression with a common ratio \(a\) converges if \(|a|<1\) and its sum \(S_{n}=\frac{1}{1-a}\), and diverges if \(|a| \geqslant 1\).