Series: Introduction

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Given a sequence \(\{a_n\}\), the expression

\[\sum_{n=0}^{\infty}a_n = a_0 + a_1 + a_2 + \cdots\]

is called a series (or infinite series). This expression may or may not have value. At this point, it is purely formal. Note that the order of addition matters: We first add \(a_0\) and \(a_1\), to the result of which we add \(a_2\), to the result of which we add \(a_3\), and so on (Not something like we first add \(a_{101}\) and \(a_{58}\), then add \(a_{333051}\), and so on). We will see, however, that for a special class of series (the positive term series), the order of addition does not matter if the series converges.



Example. The sum of a geometric progression \(\{ar^n\}\), that is, \(\sum_{n=0}^{\infty}ar^n\) is called a geometric series. It is understood that \(r^0 = 1\) including the case when \(r = 0\). □

Given a series \(\sum_{n=0}^{\infty}a_n\) and a number \(n\geq 0\), the sum

\[\sum_{k=0}^{n}a_k = a_0 + a_1 + \cdots + a_n\]

is called the \(n\)-th partial sum. We may define a new sequence \(\{S_n\}\), that is,

\[\begin{eqnarray*} S_0 &=& a_0,\\ S_1 &=& a_0 + a_1,\\ S_2 &=& a_0 + a_1 + a_2,\\ &\vdots& \end{eqnarray*}\]

This sequence \(\{S_n\}\) is called the partial sum sequence of the series \(\sum_{n=0}^{\infty}a_n\). If the partial sum sequence \(\{S_n\}\) converges, the series \(\sum_{n=0}^{\infty}a_n\) is said to have a sum or to converge. In this case, if \(\lim_{n\to\infty}S_n = \alpha\), then the value of the series \(\sum_{n=0}^{\infty}a_n\) is denoted as \(\alpha\), that is,

\[\sum_{n=0}^{\infty}a_n = \alpha.\]

Example. Consider the geometric series \(\sum_{n=0}^{\infty}ar^n\). We have

\[S_n = \sum_{k=0}^{n}ar^k = \left\{ \begin{array}{cc} \frac{a(1 - r^{n+1})}{1 - r} & (r\neq 1),\\ a(n+1) & (r=1). \end{array} \right.\]

Thus, the series has a sum when \(|r| < 1\), and then the sum is

\[\sum_{n=0}^{\infty}ar^{n} = \frac{a}{1 - r}.\]

□

Theorem (Linearity of sums)

Suppose the series \(\sum_{n=0}^{\infty}a_n\) and \(\sum_{n=0}^{\infty}b_n\) have sums. For any constants \(k\) and \(l\), \(\sum_{n=0}^{\infty}(ka_n + lb_n)\) has a sum and

\[\sum_{n=0}^{\infty}(ka_n + lb_n) = k\sum_{n=0}^{\infty}a_n + l\sum_{n=0}^{\infty}b_n.\]

Proof. Recall the linearity of limits: \(\lim_{n\to\infty}(ka_n + lb_n) = k\lim_{n\to\infty}a_n + l\lim_{n\to\infty}b_n\). The rest is exercise. ■

It should be also clear that the series \(\sum_{n=0}^{\infty}a_n\) has a sum if and only if its partial sum sequence \(\{S_n\}\) is a Cauchy sequence, that is,

  • For any \(\varepsilon > 0\), there exists an \(N\in \mathbb{N}\) such that, for all \(k, l \in \mathbb{N}\), if \(k > l \geq N\), then \(|S_k - S_l| = |a_{l+1} + a_{l+2} + \cdots + a_{k}|< \varepsilon\).

In the above condition, if \(k = l + 1 = n \geq N\), then \(|a_n| < \varepsilon\), which implies that the sequence \(\{a_n\}\) converges to 0. Thus, we have just proved the following theorem.

Theorem (Necessary condition for convergence of series)

If the series \(\sum_{n=0}^{\infty}a_n\) has a sum, then the sequence \(\{a_n\}\) converges to 0.

However, the converse of this theorem does not necessarily hold, as the next example shows.

Example. Consider the harmonic sequence \(a_n = \frac{1}{n} ~ (n = 1, 2, \cdots)\). Clearly, \(\lim_{n\to \infty}a_n = 0\). On the other hand,

\[\begin{eqnarray*} \sum_{n=1}^{\infty}\frac{1}{n} &=& 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\\ &=&\int_1^21dx + \int_2^3\frac{1}{2}dx + \int_3^4\frac{1}{3}dx + \int_4^5\frac{1}{4}dx + \cdots\\ &\geq&\int_1^2\frac{1}{x}dx + \int_2^3\frac{1}{x}dx + \int_3^4\frac{1}{x}dx + \int_4^5\frac{1}{x}dx + \cdots\\ &=& \int_1^\infty\frac{1}{x}dx\\ &=& \left[\log x\right]_1^{\infty} = \infty. \end{eqnarray*}\]

Thus, the series \(\sum_{n=1}^{\infty}\frac{1}{n}\) diverges to \(+\infty\). □

Positive term series

In a finite sum, the result of addition does not depend on the order of addition. In other words, the commutative law holds: \(a + b = b + a\).

In a series involving a sum of infinitely many terms, the commutative law may not apply (i.e., the result may depend on the order of addition).

To see why we first consider positive term series.

Definition (Positive term series)

The series \(\sum_{n=0}^{\infty}a_n\) is said to be a positive term series if \(a_n \geq 0\) for all \(n \in \mathbb{N}_0\).

Remark. \(\mathbb{N}_0 = \mathbb{N} \cup \{0\}\). □

The partial sum sequence \(\{S_n\}\) of a positive term series \(\sum_{n=0}^{\infty}a_n\) is clearly monotone increasing. If \(\{S_n\}\) is bounded above, then the series has a sum, according to the monotone convergence theorem.

Example. The series \(\sum_{n=1}^{\infty}\frac{1}{n^\alpha}\) has a sum if \(\alpha > 1\). This can be shown as follows. Let \(N\) be any natural number. If \(\alpha > 1\), then

\[\begin{eqnarray*} \sum_{n=1}^{N}\frac{1}{n^\alpha} &=& 1 + \frac{1}{2^\alpha} + \frac{1}{3^\alpha} + \cdots + \frac{1}{N^\alpha}\\ &=& 1 + \int_1^2\frac{dx}{2^\alpha} + \int_2^3\frac{dx}{3^\alpha} + \cdots + \int_{N-1}^{N}\frac{dx}{N^\alpha}\\ &\leq& 1 + \int_1^{N}\frac{dx}{x^\alpha}\\ &\leq& 1 + \int_1^{\infty}\frac{dx}{x^\alpha}\\ &=& 1 + \frac{1}{\alpha - 1} = \frac{\alpha}{\alpha -1}. \end{eqnarray*}\]

Therefore, the series is bounded above by \(\frac{\alpha}{\alpha -1}\), and hence it has a sum. □

See also: the Riemann zeta function. 

Theorem (Converging positive term series can be summed in any order)

Suppose the positive term series \(\sum_{n=0}^{\infty}a_n\) has a sum. Let us arbitrarily permute the index of the sequence \(\{a_n\}\) to define another sequence \(\{a_{n(k)}\}\). Then the series \(\sum_{k=0}^{\infty}a_{n(k)}\) also has a sum, and its value is the same as the original sum. That is,

\[\sum_{k=0}^{\infty}a_{n(k)} = \sum_{n=0}^{\infty}a_n.\]

Proof. Let \(\alpha = \sum_{n=0}^{\infty}a_n\). For each \(i = 0, 1, 2, \cdots\), consider the first \((i+1)\) terms in the series \(\sum_{k=0}^{\infty}a_{n(k)}\): \(a_{n(0)}, a_{n(1)}, a_{n(2)}, \cdots, a_{n(i)}\). Let us define \(N = \max\{n(0), n(1), \cdots, n(i)\}\). Then, all of \(a_{n(0)}, a_{n(1)}, a_{n(2)}, \cdots, a_{n(i)}\) are present among \(a_0, a_1, \cdots, a_N\) so that

\[a_{n(0)} + a_{n(1)} + a_{n(2)} + \cdots + a_{n(i)} \leq a_{0} + a_{1} + \cdots + a_N \leq \alpha.\]

This implies that the series \(\sum_{k=0}^{\infty}a_{n(k)}\) is bounded above (by \(\alpha\)), and hence it has a sum. Let \(\beta = \sum_{k=0}^{\infty}a_{n(k)}\). Then \(\beta \leq \alpha\).

Note that the terms in the original series \(\sum_{n=0}^{\infty}a_n\) may be regarded as a permutation of the terms in the series \(\sum_{k=0}^{\infty}a_{n(k)}\). Therefore, applying the same argument as above, it follows that \(\alpha \leq \beta\). Thus, \(\alpha = \beta\). ■

Definition (Dominating series)

Let \(\sum_{n=0}^{\infty}a_n\) be a positive term series. Another series \(\sum_{n=0}^{\infty}b_n\) is said to be a dominating series of \(\sum_{n=0}^{\infty}a_n\) if it has a sum and \(a_n \leq b_n\) for all \(n = 0, 1, \cdots\).

Theorem (Dominated series theorem)

A positive term series has a sum if it has a dominating series.  

Proof. Consider a positive term series \(\sum_{n=0}^{\infty}a_n\) that is dominated by the converging series \(\sum_{n=0}^{\infty}b_n = \beta\). Then, the partial sum sequence of \(\sum_{n=0}^{\infty}a_n\) is bounded above by \(\beta\). By the monotone converging theorem, the series \(\sum_{n=0}^{\infty}a_n\) converges. □



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