Convergence of series

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Absolute convergence

As we have seen in a previous post, if a positive term series has a sum, then the sum does not depend on the order of addition. However, it is not necessarily the case for general series. If a series converges absolutely, then it has some nice properties similar to positive term series.



See also: Series: Introduction

Definition (Absolute convergence, conditional convergence)

The series \(\sum_{n=0}^{\infty}a_n\) is said to converge absolutely if the series \(\sum_{n=0}^{\infty}|a_n|\) has a sum. A series is said to converge conditionally if it has a sum but does not converge absolutely.

Remark. In other words, the series \(\sum_{n=0}^{\infty}a_n\) converges conditionally if \(\sum_{n=0}^{\infty}a_n\) converges and \(\sum_{n=0}^{\infty}|a_n|\) diverges to \(+\infty\). □

Theorem (Absolutely converging series has a unique sum)

Suppose that the series \(\sum_{n=0}^{\infty}a_n\) converges absolutely. Then, the following hold:

  1. The series \(\sum_{n=0}^{\infty}a_n\) has a sum.
  2. For any arbitrarily permuted sequence of \(\{a_n\}\), say \(\{a_{n(k)}\}\), its sum is equal to the sum of the original series. That is, \[\sum_{k=0}^{\infty}a_{n(k)} = \sum_{n=0}^{\infty}a_n.\]
Proof. We define the following two sequences \(\{a_n^{+}\}\) and \(\{a_n^{-}\}\).\[\begin{eqnarray*} a_n^{+} &=& \left\{ \begin{array}{cc} a_n & \text{(if $a_n \geq 0$)},\\ 0 & \text{(if $a_n < 0$)}, \end{array}\right.\\ a_n^{-} &=& \left\{ \begin{array}{cc} 0 & \text{(if $a_n > 0$)},\\ -a_n & \text{(if $a_n\leq 0$)}. \end{array}\right. \end{eqnarray*}\] 
  1. Note that \(a_n = a_n^{+} - a_n^{-}\) for all \(n \in \mathbb{N}_0\). Both \(\sum_{n=0}^{\infty}a_n^{+}\) and \(\sum_{n=0}^{\infty}a_n^{-}\) are positive term series and are dominated by \(\sum_{n=0}^{\infty}|a_n|\). By the Dominated Series Theorem(*), both \(\sum_{n=0}^{\infty}a_n^{+}\) and \(\sum_{n=0}^{\infty}a_n^{-}\) converge. By the linearity of sums(*), \(\sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(a_n^{+}-a_n^{-})\) also converges.
  2. Both \(\sum_{n=0}^{\infty}a_n^{+}\) and \(\sum_{n=0}^{\infty}a_n^{-}\) are positive term series so that their permuted series converge to the same values. Therefore, any permuted series of \(\sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(a_n^{+}-a_n^{-})\) converges to the same value.
(*) For the Dominated Series Theorem and the linearity of sums, see Series: Introduction. ■

On the contrary, we have the following, perhaps mind-boggling, theorem for conditionally convergent series (proof is omitted).

Theorem (Conditionally converging series, when permuted, can converge to arbitrary values)

Suppose \(\sum_{n=0}^{\infty}a_n\) converges conditionally. Then, for any \(\alpha \in \mathbb{R}\), there exists a permuted sequence \(\{a_{n(k)}\}\) of the sequence \(\{a_n\}\) such that
\[\sum_{k=0}^{\infty}a_{n(k)} = \alpha.\]
Proof. Omitted. ■

Convergence criteria

We have a few handy theorems to judge if a series converges.

Theorem (Cauchy's criterion)

Let \(\sum_{n=0}^{\infty}a_n\) be a positive term series. Suppose the limit \(\lim_{n\to\infty}\sqrt[n]{a_n} = r\) exists.
  1. If \(r < 1\), then the series \(\sum_{n=0}^{\infty}a_n\) converges.
  2. If \(r > 1\), then the series \(\sum_{n=0}^{\infty}a_n\) diverges.
Proof. 
  1. Suppose \(r < 1\). Choose a \(t\in\mathbb{R}\) such that \(r < t < 1\). Then, \(\sqrt[n]{a_n} \leq t\) holds for all but finitely many \(n\). In fact, if we set \(\varepsilon = t - r > 0\), there exists some \(N\in\mathbb{N}\) such that if \(n \geq N\) then \(|\sqrt[n]{a_n} - r| < \varepsilon\), and hence, in particular, \(\sqrt[n]{a_n} < r + \varepsilon = t\). In this case, we have \(a_n \leq t^n\) for all but finitely many \(n\). Recall that, since \(0 < t < 1\), \(\sum_{n=0}^{\infty}t^n = 1/(1 - t)\). Therefore, by the Dominated Series Theorem, the series \(\sum_{n=0}^{\infty}a_n\) converges.
  2. If \(r > 1\), \(\sqrt[n]{a_n} \geq 1\) for all but finitely many \(n\). Thus, the sequence \(\{a_n\}\) does not converge to 0 as \(n\to \infty\). This means that the series does not satisfy the necessary condition for convergence (*). Therefore, the series \(\sum_{n=0}^{\infty}a_n\) diverges.
(*) For the necessary condition for the converging series mentioned above, see Series: Introduction. ■

Remark. From the above proof, we can see the following:
  • If there exists an \(r < 1\) such that \(\sqrt[n]{a_n} \leq r\) for all but finitely many \(n\), then the series \(\sum_{n=0}^{\infty}a_n\) converges.

□

Example. The positive term series \(\sum_{n=0}^{\infty}\left(\frac{an + b}{cn + d}\right)^n ~ (a, b, c, d > 0)\) converges if \(a < c\). In fact, 

\[\lim_{n \to \infty}\sqrt[n]{\left(\frac{an + b}{cn + d}\right)^n} = \lim_{n \to \infty}\frac{an + b}{cn + d} = \frac{a}{c}.\]

Thus, if \(\frac{a}{c} < 1\), then the given series converges. □

Theorem (D'Alembert's criterion)

Let \(\sum_{n=0}^{\infty}a_n\) be a positive term series. Suppose that the limit \(\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}} = r\) exists.

  1. If \(r < 1\), then the series \(\sum_{n=0}^{\infty}a_n\) converges.
  2. If \(r > 1\), then the series \(\sum_{n=0}^{\infty}a_n\) diverges.
Proof. 
  1. Suppose \(r < 1\). We can choose a real number \(t\) such that \(r < t < 1\). Then, \(\frac{a_{n+1}}{a_n} \leq t\) holds for all but finitely many \(n\). Thus, for a sufficiently large \(N\), if \(n \geq N\), then \[a_n \leq ta_{n-1} \leq t^2 a_{n-2} \leq \cdots \leq t^{n-N}a_N.\] The (dominating) series \(\sum_{n=N}^{\infty}t^{n-N}a_{N} = a_N\sum_{n=0}^{\infty}t^{n}\) converges. Therefore the series \(\sum_{n=0}^{\infty}a_n\) converges.
  2. If \(r > 1\), \(a_{n+1} \geq a_n\) for all but finitely many \(n\). Thus, \(\lim_{n\to\infty}a_n > 0\). Thus, the given series diverges.
■

Example. The series \(\sum_{n=0}^{\infty}\frac{a^n}{n!} ~ (a > 0)\) converges. In fact, if we set \(a_n = \frac{a^n}{n!}\),
\[\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^{n}}{n!}} = \lim_{n\to\infty}\frac{a}{n+1} = 0 < 1.\]
By D'Alembert's criterion, the series has a sum. □

Remark. What happens when \(\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1\)? D'Alembert's criterion is useless in this case. For example, \(\sum_{n=1}^{\infty}\frac{1}{n}\) diverges whereas \(\sum_{n=1}^{\infty}\frac{1}{n^2}\) converges, but both satisfy \(\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1\). □


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