Infinite Series

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

The properties of infinite sequences discussed in §Limits of Sequences are particularly important when we deal with infinite series.

Partial Sums and Infinite Series

Definition 1 If the sequence of partial sums \(s_N = \sum_{n=1}^N a_n\) of a sequence \((a_n)\) converges to \(S\), we say that the infinite series \(\sum a_n\) converges to \(S\) and write \(\sum a_n = S\). If \((s_N)\) diverges, we say the series diverges.

Thus, the convergence or divergence of a series reduces to that of the sequence of partial sums \(s_N\), so we may apply the tools from §Limits of Sequences directly. The following are partial sums of two particularly useful series.

Example 2

• The geometric series \(\sum r^n\) is the infinite series defined by the partial sum

  \[s_N = \sum_{n=0}^N r^n = \frac{1 - r^{N+1}}{1-r} \qquad (r \neq 1)\]

  which is verified by \((1-r)s_N = (1-r)(1 + r + \cdots + r^N) = 1 - r^{N+1}\). If \(\lvert r\rvert < 1\), then \(r^{N+1} \rightarrow 0\), so the infinite sum converges to \(1/(1-r)\). If \(\lvert r\rvert \geq 1\), the sequence is not bounded below or oscillates, hence diverges.

• The \(p\)-series \(\sum_n 1/n^p\) converges if \(p > 1\) and diverges if \(p \leq 1\). If \(p \leq 0\), the general term \(1/n^p = n^{-p}\) diverges to infinity, so in particular the partial sums are not bounded above and thus diverge. Consider the case \(p > 0\). Excluding the first term of the partial sum and grouping the remaining terms in blocks of \(2^k\) \((k=1,\dots,m)\), we examine up to the \((2^{m+1}-1)\)-st term:

  \[s_{2^{m+1}-1} = 1 + \left(\frac{1}{2^p}+\frac{1}{3^p}\right) + \left(\frac{1}{4^p}+\cdots+\frac{1}{7^p}\right) + \cdots + \left(\frac{1}{(2^m)^p}+\cdots+\frac{1}{(2^{m+1}-1)^p}\right)\]

  Here, the \(k\)-th parenthesis contains \(2^k\) terms, and every term inside is at most \(1/(2^k)^p\), so

  \[s_{2^{m+1}-1} \leq \sum_{k=0}^m 2^k \cdot \frac{1}{2^{kp}} = \sum_{k=0}^m (2^{1-p})^k\]

  If \(p > 1\), then \(2^{1-p} < 1\), so the right-hand side is a partial sum of a geometric series and is bounded; applying §Limits of Sequences, ⁋Proposition 7 (Monotone Convergence), we conclude that the \(p\)-series converges. On the other hand, if \(0 < p \leq 1\), the sum inside each parenthesis is at least the smallest term \(1/(2^{k+1}-1)^p\) among the \(2^k\) terms, so

  \[\frac{2^k}{(2^{k+1}-1)^p} \geq \frac{2^k}{2^{k+1}-1} \geq \frac{1}{2}\]

  Therefore,

  \[s_{2^{m+1}-1} \geq 1 + \frac{1}{2} + \frac{1}{2} + \cdots + \frac{1}{2} \geq 1 + \frac{m}{2}\]

  and we confirm that the partial sums diverge to infinity.

The perspective of Definition 1—that a series reduces to the convergence problem of its sequence of partial sums—implies that properties of limits of sequences carry over directly to series. The most basic is linearity: the termwise sum and scalar multiple of two convergent series again preserve convergence. This follows from the linearity of limits at the level of partial sums (§Limits of Sequences, ⁋Proposition 2 (Limit Laws for Sequences)).

Proposition 3 (Linearity of Series) If \(\sum a_n = A\) and \(\sum b_n = B\) converge and \(\lambda, \mu\) are constants, then \(\sum (\lambda a_n + \mu b_n)\) also converges and its sum is \(\lambda A + \mu B\).

Proof

Let the partial sums be \(s_N = \sum_{n=1}^N a_n\) and \(t_N = \sum_{n=1}^N b_n\). Since these are finite sums,

\[\begin{aligned} \sum_{n=1}^N (\lambda a_n + \mu b_n) &= \lambda \sum_{n=1}^N a_n + \mu \sum_{n=1}^N b_n \\ &= \lambda s_N + \mu t_N \end{aligned}\]

Since \(s_N \rightarrow A\) and \(t_N \rightarrow B\), by linearity of convergent sequences the right-hand side converges to \(\lambda A + \mu B\). Thus the partial sum on the left-hand side also converges to the same value, yielding the conclusion.

Convergence Tests

For most series, finding the exact sum is too ambitious; often only the convergence behavior is needed. The most basic yet powerful tool for determining this is the following proposition.

Proposition 4 (Term Test) If the series \(\sum a_n\) converges, then \(a_n \rightarrow 0\). Therefore, if \(a_n \not\rightarrow 0\), the series diverges.

Proof

If \(s_N \rightarrow S\), then also \(s_{N-1} \rightarrow S\), so \(a_N = s_N - s_{N-1} \rightarrow S - S = 0\).

Example 5 (Harmonic Series) The harmonic series

\[\sum \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots\]

has terms tending to \(0\) yet diverges. This is the case \(p=1\) of Example 2; grouping the terms in blocks of \(2^k\),

\[1 + \frac{1}{2} + \left(\frac{1}{3}+\frac{1}{4}\right) + \left(\frac{1}{5}+\cdots+\frac{1}{8}\right) + \cdots\]

we see that the sum of each block is at least \(1/2\), so the partial sums grow without bound.

As the harmonic series shows, \(a_n \rightarrow 0\) is a necessary but not sufficient condition for convergence. On the other hand, for a series with positive terms the partial sums are monotone increasing, so by §Limits of Sequences, ⁋Proposition 7 (Monotone Convergence) the convergence question reduces to checking boundedness.

Theorem 6 For a series with positive terms, the following hold.

1. If \(0 \leq a_n \leq b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
2. (Ratio Test) If \(a_{n+1}/a_n \rightarrow L\), then the series converges when \(L < 1\) and diverges when \(L > 1\).

Proof

1: The partial sums \((s_N)\) of \(\sum a_n\) are increasing and bounded above by \(\sum_n b_n\), so by the monotone convergence theorem (§Limits of Sequences, ⁋Proposition 7 (Monotone Convergence)) they converge.

2: If \(L < 1\), choose \(r\) with \(L < r < 1\); then for sufficiently large \(n\) we have \(a_{n+1} \leq r a_n\), hence \(a_{N+k} \leq r^k a_N\), and the series is compared with a convergent geometric series. If \(L > 1\) the terms increase and do not tend to \(0\), so the series diverges by the term test.

Observe that in the ratio test the case \(L = 1\) is inconclusive. For instance, all \(p\)-series have \(L = 1\), yet their convergence behavior depends on \(p\); for this reason, when testing convergence of a series it is important to choose the appropriate test.

The comparison test is among the most powerful of the tests examined in this article, since it can be applied to any series as long as a suitable inequality is found, but it carries the burden of having to bound the terms directly by inequalities with another series. In practice, the terms of a series are often clearly of the same order of magnitude as some standard series, yet a clean inequality is hard to obtain. In this case the following variant is convenient, as it requires only the limit of the ratio of terms.

Proposition 7 (Limit Comparison Test) For positive-term series \(\sum a_n\) and \(\sum b_n\), if \(a_n/b_n \rightarrow c\) with \(0 < c < \infty\), then the two series converge or diverge together.

Proof

From the definition of the limit, for sufficiently large \(n\) we can arrange

\[\frac{c}{2} < \frac{a_n}{b_n} < 2c\]

Multiplying by \(b_n > 0\),

\[\begin{aligned} \frac{c}{2} b_n &< a_n < 2c b_n \end{aligned}\]

If \(\sum b_n\) converges, then \(\sum 2c b_n\) also converges, so by the right-hand inequality and the comparison test (Theorem 6), \(\sum a_n\) converges. Conversely, if \(\sum a_n\) converges, then from the left-hand inequality \(b_n < (2/c) a_n\) and the comparison test, \(\sum b_n\) converges. The divergence case follows by contrapositive.

The utility of the limit comparison test lies in replacing terms involving mixed polynomials and powers by the dominant term whose convergence behavior governs the series. If the term is a fraction, we need only retain the fastest-growing terms in the numerator and denominator as a power, take that as the standard \(b_n\), and check whether the limit of \(a_n/b_n\) lies in \((0, \infty)\), eliminating the labor of manually matching inequalities.

Another effective test is the following root test.

Proposition 8 (Root Test) For a positive-term series \(\sum a_n\), if \(\sqrt[n]{a_n} \rightarrow L\), then the series converges when \(L < 1\) and diverges when \(L > 1\).

Proof

If \(L < 1\), we can choose \(r\) with \(L < r < 1\), and for sufficiently large \(n\) we have \(\sqrt[n]{a_n} < r\), i.e. \(a_n < r^n\). The right-hand side is a convergent geometric series with ratio \(r < 1\), so by the comparison test \(\sum a_n\) converges. If \(L > 1\), then for sufficiently large \(n\) we have \(a_n > 1\), so the terms do not tend to \(0\) and the series diverges by the term test (Proposition 4 (Term Test)).

In general, the comparison tests may appear more convenient to apply to sequences, but the root test can decide convergence in more cases. It is particularly powerful for series whose terms are given in the form of an \(n\)-th power; for example, in \(\sum (n/(2n+1))^n\) we immediately read off \(\sqrt[n]{a_n} = n/(2n+1) \rightarrow 1/2 < 1\), so convergence follows. However, the root test also gives no conclusion when \(L = 1\), just like the ratio test, so in this case another method must be tried.

Absolute Convergence and Alternating Series

Positive-term series are easy to test for convergence because their partial sums are monotone increasing, relying on §Limits of Sequences, ⁋Proposition 7 (Monotone Convergence) and tools including Theorem 6; on the other hand, they are harder to make converge than series with negative terms mixed in. In this section we examine series whose signs alternate.

Definition 9 If \(\sum \lvert a_n\rvert\) converges, we say that \(\sum a_n\) converges absolutely.

An absolutely convergent series always converges. Since \(0 \leq a_n + \lvert a_n\rvert \leq 2\lvert a_n\rvert\), by Theorem 6 the series \(\sum (a_n + \lvert a_n\rvert)\) converges, and therefore \(\sum a_n = \sum (a_n + \lvert a_n\rvert) - \sum \lvert a_n\rvert\) also converges. However, the converse does not hold. When positive and negative terms are mixed, the partial sums can cancel each other, so the original series may converge even though the series of absolute values diverges. The following test is the most basic tool guaranteeing such convergence.

Proposition 10 (Alternating Series Test) If a sequence of positive numbers \(b_n \geq 0\) is monotone decreasing and \(b_n \rightarrow 0\), then the alternating series \(\sum (-1)^{n+1} b_n\) constructed from it converges. Moreover, the error between its sum \(S\) and the partial sum \(s_N\) is controlled by \(\lvert S - s_N\rvert \leq b_{N+1}\).

Proof

The even partial sums

\[s_{2m} = (b_1 - b_2) + \cdots + (b_{2m-1} - b_{2m})\]

are increasing since each parenthesis is nonnegative, and

\[s_{2m} = b_1 - (b_2 - b_3) - \cdots - b_{2m} \leq b_1\]

so they are bounded above. Thus \(s_{2m}\) converges, and since \(s_{2m+1} = s_{2m} + b_{2m+1}\) with \(b_{2m+1} \rightarrow 0\), the odd partial sums also converge to the same limit \(S\). Since the partial sums alternate around \(S\), the error claim also follows.

A series that converges but not absolutely is called conditionally convergent. For example, the alternating harmonic series \(\sum (-1)^{n+1}/n = \ln 2\) converges by the alternating series test, but taking absolute values yields the harmonic series which diverges, so it is conditionally convergent.

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.