Proposition 3.1.1. Divergence Test.
If the series \(\sum x_n\) converges, then \(x_n\to 0\text{.}\)
Section 3.1 Series
A series is the sum of a sequence. We write
\begin{equation*}
\sum_{n=m}^k x_n
\end{equation*}
for the finite sum
\begin{equation*}
x_m+x_{m+1}+\cdots+x_k.
\end{equation*}
This is just convenient notation, but it will also guide the definition of an infinite series.
Given an infinite sequence \((x_n)\text{,}\) the expression \(\sum_{n=1}^\infty x_n\) is called an infinite series. To make sense of it, we consider its sequence of partial sums\((s_k)\text{,}\) where
\begin{equation*}
s_k:=\sum_{n=1}^k x_n = x_1+x_2+\cdots+x_k.
\end{equation*}
We say that the series \(\sum_{n=1}^\infty x_n\)converges if the sequence \((s_k)\) converges. In that case, we write
\begin{equation*}
\sum_{n=1}^\infty x_n = s
\end{equation*}
where \(s\) is the limit of \((s_k)\text{.}\) The series diverges if it does not converge. We also write \(\sum_{n=1}^\infty x_n = +\infty\) (resp. \(-\infty) \)if \((s_k)\) diverges to \(+\infty\) (resp. \(-\infty\)).
Since altering finitely many terms will not affect the convergence of a series, so if we concern only about convergence, we will drop the indices and simply write \(\sum x_n\) for the series whose terms are from the sequence \((x_n)\text{.}\)
Proof.
Suppose \(\sum x_n\) converges to \(s\text{.}\) That mean it sequence of partial sums \((s_n)\) converges to \(s\text{.}\) But then the sequence \((s_{n+1})\) also has \(s\) as its limit. Therefore,
\begin{equation*}
x_{n+1} = s_{n+1} - s_{n} \to 0.
\end{equation*}
This show that \((x_n)\) is null.
Example 3.1.2.
The series \(\sum_{n=1}^\infty 1\) and \(\sum_{n=1}^{\infty}
(-1)^{n+1}\) are both divergent by the divergence test. These series are examples of geometry series.
A geometric series is a series whose terms form a geometric series. We will show in the next section that the geometric series \(\sum r^n\) converges when \(|r| \lt 1\) and diverges when \(|r| \ge 1\text{.}\)
The converse statement of the divergence test is false: a null sequence need not give a convergent series. The standard example is the harmonic series\(\sum_{n=1}^\infty 1/n\text{.}\) We will prove in the next section that this series diverges even though \(1/n\to 0\text{.}\)
A series \(\sum x_n\) is absolutely convergent if the series
\begin{equation*}
\sum |x_n|
\end{equation*}
converges. It is conditionally convergent if \(\sum x_n\) converges but \(\sum |x_n|\) diverges.
Example 3.1.3.
The series \(\sum 1/n^2\) is absolutely convergent, while the alternating harmonic series
\begin{equation*}
\sum \frac{(-1)^{n+1}}{n}
\end{equation*}
is conditionally convergent. We defer the justifications to the next section, where we develop tests for convergence of series.
The distinction between absolute and conditional convergence is important. Absolute convergence behaves much like finite summation, but conditional convergence can be surprisingly unstable under rearrangement of terms.
A rearrangement of the series \(\sum_{n=1}^\infty x_n\) is a series of the form \(\sum_{n=1}^\infty x_{\pi(n)}\text{,}\) where \(\pi:\N\to\N\) is a bijection.
Theorem 3.1.4. Riemann Rearrangement Theorem.
If \(\sum_{n=1}^\infty x_n\) is conditionally convergent and \(L\in\R\text{,}\) then there is a rearrangement of the series that converges to \(L\text{.}\) Moreover, there are rearrangements that diverge to \(+\infty\) and rearrangements that diverge to \(-\infty\text{.}\)
We will not prove this theorem here. A proof can be found at
ProofWiki. That page also cites Rudin, "Principles of Mathematical Analysis", Theorem 3.54.
