Infinite Series

The properties of infinite sequences discussed in §Limits of Sequences are used importantly in our context, especially when dealing with infinite series.

Partial Sums and Infinite Series

Definition 1 For a sequence \((a_n)\), if the partial sums \(s_N = \sum_{n=1}^N a_n\) form a sequence \((s_N)\) that 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, the series is said to diverge.

According to this, the convergence and divergence of a series is reduced to the problem of the convergence and divergence of the sequence \(s_N\) called the partial sums, so the tools from §Limits of Sequences can be used as they are. The following are the partial sums of two particularly frequently used series.

Example 2

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

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

  and the equation above can be verified through \((1-r)s_N = (1-r)(1 + r + \cdots + r^N) = 1 - r^{N+1}\). If \(\lvert r\rvert < 1\), then \(r^{N+1} \rightarrow 0\), so we know that this infinite sum converges to \(1/(1-r)\). If \(\lvert r\rvert \geq 1\), one can prove that this sequence diverges because it is not bounded above or oscillates.

• The \(p\)-series \(\sum_n 1/n^p\) converges if \(p > 1\) and diverges if \(p \leq 1\). If \(p \leq 0\), since the general term is \(1/n^p = n^{-p} \geq 1\) for all \(n\), we have \(s_N \geq N\); in particular, the partial sums are not bounded above and therefore diverge. Consider the case \(p > 0\). If, excluding the first term of the partial sum, we group terms into blocks of \(2^k\) \((k=1,\dots,m)\) and look up to the \(2^{m+1}-1\)st term, we have

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

  Here, the \(k\)th bracket contains \(2^k\) terms, and every term inside it is less than or equal to \(1/(2^k)^p\), 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 therefore bounded; applying §Limits of Sequences, ⁋Proposition 7 (Monotone convergence), one can show that the \(p\)-series converges. On the other hand, if \(0 < p \leq 1\), the sum of each bracket is at least as large as the smallest term \(1/(2^{k+1}-1)^p\) among the \(2^k\) terms, so it is at least

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

  Therefore, from

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

  we can confirm that the partial sums diverge to infinity.

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

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

If we let the partial sums be \(s_N = \sum_{n=1}^N a_n\) and \(t_N = \sum_{n=1}^N b_n\), respectively, then since these are finite sums,

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

holds. Since \(s_N \rightarrow A\) and \(t_N \rightarrow B\), the linearity of convergent sequences implies that the right-hand side converges to \(\lambda A + \mu B\). Hence the partial sums on the left also converge to the same value, giving the conclusion.

Convergence Tests

For most series, finding the sum is too ambitious, and in many cases only whether it converges is needed. The most basic yet powerful tool for determining this is the following proposition.

Proposition 4 (General Term Test) If the series \(\sum a_n\) converges, then \(a_n \rightarrow 0\). Hence, 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 \(0\) but diverges. This is the special case \(p=1\) of Example 2; similarly, if we group the terms in blocks of \(2^k\) and look at

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

the sum in each pair of parentheses is at least \(1/2\), so the partial sums grow without bound.

As the harmonic series shows, \(a_n \rightarrow 0\) is only a necessary condition for convergence, not a sufficient one. 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), convergence reduces to checking whether the sum is bounded above.

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

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

Proof

1: For \(n > N_0\), suppose \(0 \leq a_n \leq b_n\), let the partial sums of \(\sum b_n\) be \(t_N\), and let their limit be \(B\). Since \(b_n \geq 0\), we have \(t_N \leq B\), and thus for \(N > N_0\),

\[s_N = s_{N_0} + \sum_{n=N_0+1}^N a_n \leq s_{N_0} + \sum_{n=N_0+1}^N b_n = s_{N_0} + t_N - t_{N_0} \leq s_{N_0} + B - t_{N_0}\]

The right-hand side is independent of \(N\), so simply adding the correction from the finitely many earlier terms shows that the partial sums of \(\sum a_n\), \((s_N)\), are bounded above. Since \((s_N)\) is increasing, it converges by §Limits of Sequences, ⁋Proposition 7 (Monotone convergence).

2: If \(L < 1\), we can choose \(L < r < 1\); then for such an \(r\), for large \(n\) we have \(a_{n+1} \leq r a_n\), so \(a_{N+k} \leq r^k a_N\), and comparison with a convergent geometric series applies. If \(L > 1\), the terms increase and do not tend to \(0\), so the series diverges by Proposition 4 (General Term Test).

Observe that the ratio test is inconclusive when \(L = 1\). For instance, all \(p\)-series have \(L = 1\), yet their convergence depends on \(p\); for this reason, choosing an appropriate test is important when determining the convergence of a series.

The comparison test can be applied to any sequence, no matter what it looks like, provided a suitable inequality can be found, so it is among the most powerful tests we examine in this post; however, it carries the burden of having to bound the terms directly with those of another series via an inequality. In actual series, it is often clear that the terms are of roughly the same size as those of some reference series, yet a clean inequality is often elusive. In such cases, the following variant is convenient: it suffices to examine only the limit of the ratio between terms.

Proposition 7 (Limit Comparison Test) For series of positive terms \(\sum a_n\) and \(\sum b_n\), if \(a_n/b_n \rightarrow c\) and \(0 < c < \infty\), then the two series either both converge or both diverge.

Proof

From the definition of the limit, for all sufficiently large \(n\) we can ensure that

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

holds. Multiplying both sides 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 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 case of divergence follows by contraposition.

The utility of the limit comparison test lies in replacing a term involving mixed polynomials and powers with the term that dominates its convergence behavior. If the terms are of fractional form, one can take as the reference \(b_n\) the power obtained by keeping only the fastest-growing terms in the numerator and denominator, and check whether the limit of \(a_n/b_n\) lies in \((0, \infty)\), eliminating the labor of matching inequalities by hand.

Another effective test is the following root test.

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

Proof

If \(L < 1\), we can choose \(L < r < 1\); for such \(r\), for sufficiently large \(n\) we have \(\sqrt[n]{a_n} < r\), that is, \(a_n<r^n\). 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 Proposition 4 (General Term Test).

In general, the ratio test may seem more convenient to apply to sequences, but the root test can decide convergence in more cases. It is especially 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, like the ratio test, the root test gives no conclusion when \(L=1\), and other methods must be tried in that case.

Absolute Convergence and Alternating Series

While series of positive terms have monotone increasing partial sums, making convergence easy to determine via §Limits of Sequences, ⁋Proposition 7 (Monotone convergence) and tools such as Theorem 6, they are harder to converge than series with mixed signs. In this section we examine series whose terms alternate in sign.

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

An absolutely convergent series always converges. Indeed, since \(0 \leq a_n + \lvert a_n\rvert \leq 2\lvert a_n\rvert\), Theorem 6 implies that \(\sum (a_n + \lvert a_n\rvert)\) 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 terms \(b_n \geq 0\) 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. Hence \(s_{2m}\) converges, and since \(s_{2m+1} = s_{2m} + b_{2m+1}\) with \(b_{2m+1} \rightarrow 0\), the odd partial sums converge to the same limit \(S\). On the other hand, \(s_{2m+1} - s_{2m-1} = -b_{2m} + b_{2m+1} \leq 0\), so the odd partial sums are decreasing; thus \(s_{2m} \leq S \leq s_{2m+1}\), and the partial sums straddle \(S\) alternately, yielding the error estimate as well.

A series that converges but does not converge absolutely is called conditionally convergent. For example, the alternating harmonic series \(\sum (-1)^{n+1}/n\) converges by the alternating series test, but taking absolute values gives the harmonic series, which diverges; hence 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.