Convergence of series

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 $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ is said to converge absolutely if the series $\sum _{n=0}^{\mathrm{\infty }}|a_n|$ 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 $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges conditionally if $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges and $\sum _{n=0}^{\mathrm{\infty }}|a_n|$ diverges to $+\mathrm{\infty }$. □

Theorem (Absolutely converging series has a unique sum)

Suppose that the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges absolutely. Then, the following hold:

1. The series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ 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, $$\sum _{k=0}^{\mathrm{\infty }}{a}_{n(k)}=\sum _{n=0}^{\mathrm{\infty }}{a}_n.$$

Proof. We define the following two sequences $\{a_n^{+}\}$ and $\{a_n^{-}\}$.$$\begin{array}{rcl}{a}_n^{+}& =& \{\begin{array}{cc}{a}_n& \text{(if }{a}_n\ge 0\text{)},\\ 0& \text{(if }{a}_n<0\text{)},\end{array}\\ a_n^{-}& =& \{\begin{array}{cc}0& \text{(if }{a}_n>0\text{)},\\ -a_n& \text{(if }{a}_n\le 0\text{)}.\end{array}\end{array}$$ 

1. Note that $a_n=a_n^{+}-a_n^{-}$ for all $n\in {\mathbb{N}}_0$. Both $\sum _{n=0}^{\mathrm{\infty }}{a}_n^{+}$ and $\sum _{n=0}^{\mathrm{\infty }}{a}_n^{-}$ are positive term series and are dominated by $\sum _{n=0}^{\mathrm{\infty }}|a_n|$. By the Dominated Series Theorem(*), both $\sum _{n=0}^{\mathrm{\infty }}{a}_n^{+}$ and $\sum _{n=0}^{\mathrm{\infty }}{a}_n^{-}$ converge. By the linearity of sums(*), $\sum _{n=0}^{\mathrm{\infty }}{a}_n=\sum _{n=0}^{\mathrm{\infty }}(a_n^{+}-a_n^{-})$ also converges.
2. Both $\sum _{n=0}^{\mathrm{\infty }}{a}_n^{+}$ and $\sum _{n=0}^{\mathrm{\infty }}{a}_n^{-}$ are positive term series so that their permuted series converge to the same values. Therefore, any permuted series of $\sum _{n=0}^{\mathrm{\infty }}{a}_n=\sum _{n=0}^{\mathrm{\infty }}(a_n^{+}-a_n^{-})$ 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 $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges conditionally. Then, for any $\alpha \in \mathbb{R}$, there exists a permuted sequence $\{a_{n(k)}\}$ of the sequence $\{a_n\}$ such that

$$\sum _{k=0}^{\mathrm{\infty }}{a}_{n(k)}=\alpha .$$

Proof. Omitted. ■

Convergence criteria

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

Theorem (Cauchy's criterion)

Let $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ be a positive term series. Suppose the limit $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{a_n}=r$ exists.

1. If $r<1$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges.
2. If $r>1$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ diverges.

Proof. 

1. Suppose $r<1$. Choose a $t\in \mathbb{R}$ such that $r<t<1$. Then, $\sqrt[n]{a_n}\le t$ holds for all but finitely many $n$. In fact, if we set $\epsilon =t-r>0$, there exists some $N\in \mathbb{N}$ such that if $n\ge N$ then $|\sqrt[n]{a_n}-r|<\epsilon$, and hence, in particular, $\sqrt[n]{a_n}<r+\epsilon =t$. In this case, we have $a_n\le t^n$ for all but finitely many $n$. Recall that, since $0<t<1$, $\sum _{n=0}^{\mathrm{\infty }}{t}^n=1/(1-t)$. Therefore, by the Dominated Series Theorem, the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges.
2. If $r>1$, $\sqrt[n]{a_n}\ge 1$ for all but finitely many $n$. Thus, the sequence $\{a_n\}$ does not converge to 0 as $n\to \mathrm{\infty }$. This means that the series does not satisfy the necessary condition for convergence (*). Therefore, the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ 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 $\sqrt[n]{a_n}\le r$ for all but finitely many $n$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges.

□

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

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

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

Theorem (D'Alembert's criterion)

Let $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ be a positive term series. Suppose that the limit $\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}=r$ exists.

1. If $r<1$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges.
2. If $r>1$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ diverges.

Proof. 

1. Suppose $r<1$. We can choose a real number $t$ such that $r<t<1$. Then, $\frac{a_{n+1}}{a_n}\le t$ holds for all but finitely many $n$. Thus, for a sufficiently large $N$, if $n\ge N$, then $$a_n\le t{a}_{n-1}\le t^2{a}_{n-2}\le \cdots \le t^{n-N}{a}_N.$$ The (dominating) series $\sum _{n=N}^{\mathrm{\infty }}{t}^{n-N}{a}_N=a_N\sum _{n=0}^{\mathrm{\infty }}{t}^n$ converges. Therefore the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges.
2. If $r>1$, $a_{n+1}\ge a_n$ for all but finitely many $n$. Thus, $\underset{n\to \mathrm{\infty }}{lim}{a}_n>0$. Thus, the given series diverges.

■

Example. The series $\sum _{n=0}^{\mathrm{\infty }}\frac{a^n}{n!}(a>0)$ converges. In fact, if we set $a_n=\frac{a^n}{n!}$,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}=\underset{n\to \mathrm{\infty }}{lim}\frac{a}{n+1}=0<1.$$

By D'Alembert's criterion, the series has a sum. □

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