In general, if we try to add the terms of an infinite sequence ${a_n}_{n=1}^{\infty}$ we get an expression of the form $$a_1+a_2+a_3+\cdots+a_n+\cdots$$
which is called an infinite series (or just a series) and is denoted, for short, by the symbol $$\sum_{n=1}^{\infty} a_n \\ \text{or}\\ \sum{a_n}$$
Given a series $\sum_{n=1}^{\infty}a_n = a_1 + a_2 + a_3 + \cdots$, let $s_n$ denote its nth partial sum: $$s_n = \sum_{i=0}^n a_i = a_1 + a_2 + a_3 + \cdots + a_n$$
If the sequence ${s_n}$ is convergent and $\lim_{n \to \infty} s_n = s$ exists as a real number, then the series $\sum a_n$ is called convergent and we write $$a_1 + a_2 + a_3 + \cdots + a_n + \cdots = s \\ \text{or}\\ \sum_{n=1}^{\infty} a_n = s$$
The number s is called the sum of the series. If the sequence ${s_n}$ is divergent, then the series is called divergent.
$a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} + \cdots = \sum_{n=1}^{\infty} ar^{n-1}, \\ a \ne 0$
Each term is obtained from the preceding one by multiplying it by the common ratior.
If $r = 1$, then $s_n = a + a + a + \cdots + a = na \to \pm \infty$. Since $\lim_{n \to \infty} s_n$ doesn’t exist, the geometric series diverges in this case.
If $r \ne 1$, we have $$\begin{aligned} s_n &= a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} \\ rs_n &= ar + ar^2 + ar^3 + \cdots + ar^{n}
\end{aligned}$$
Subtracting these equations, we get $$s_n - rs_n = a - ar^n$$
$$s_n = \frac{a(1-r^n)}{1-r}$$
If $-1 \le r \le 1$, then $r^n \to 0,\\ \text{as}\\ n \to \infty$, so $$\lim_{n \to \infty}s_n = \lim_{n \to \infty}\frac{a(1-r^n)}{1-r} = \frac{a}{1-r} - \frac{a}{1-r}\lim_{n \to \infty}r^n = \frac{a}{1-r}$$
Summarize the results The geometric series $$\sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + \cdots$$ is convergent if $|r| < 1$ and its sum is $$\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}\\ \\ \\ \\ |r| < 1$$ If $|r| \ge 1$, the geometric series is divergent.
Another way to get the conclusion ($-1 \le r \le 1$):
This figure provides a geometric demonstration of the result in the example. If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles, $$\frac{s}{a} = \frac{a}{a- ar}\\ \text{, so}\\ s= \frac{a}{1-r}$$
Solution: For this particular series it’s convenient to consider the partial sums $s_2, s_4, s_8, s_{16}, s_{32}, \ldots$ and show that they become large.
Similarly, $s_{32} > 1 + \frac{5}{2}, s_{64} > 1 + \frac{6}{2}$, and in general $$s_{2^n} > 1 + \frac{n}{2}$$
If the series $\displaystyle\sum_{n=1}^{\infty} a_n$ is convergent, then $\displaystyle\lim_{n \to \infty} a_n = 0$.
If $\displaystyle\lim_{n \to \infty} a_n$ does not exist or if $\displaystyle\lim_{n \to \infty} a_n \ne 0$, then the series $\displaystyle\sum_{n=1}^{\infty}a_n$ is divergent.
If $\sum a_n$ and $\sum b_n$ are convergent series, then so are the series $\sum c a_n$ (where c is a constant), $\sum(a_n + b_n)$, and $\sum(a_n - b_n)$, and
$\displaystyle\sum_{n=1}^{\infty} c a_n = c \sum_{n=1}^{\infty} a_n$
The sequence ${a_k}$ decreasing and $a_k > 0$. The series $\displaystyle\sum_{k=1}^{\infty} a_k$ converges if and only if $\displaystyle\sum_{k=0}^{\infty} 2^k \cdot a_{2^k}$ converges.
Proven:
$a_k > 0$, so $\displaystyle\sum_{k=1}^{\infty} a_k$ is non-decreasing.
Same as the Harmonic Series part, we can get: $$\begin{aligned}
s_n &= 1 + (a_2 + a_3) + (a_4 + a_5 + a_6 + a_7 + a_8) + \cdots \\ s_n &< 1 + 2 \cdot a_2 + 4 \cdot a_4 + \cdots \\ s_n &< \sum_{k=0}^{\infty} 2^k \cdot a_{2^k}
\end{aligned}$$
Same example: Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges?
