Infinite Series

The properties of infinite sequences discussed in §Limits of Sequences are especially important when we study infinite series.

Partial Sums and Infinite Series

Definition 1 If the sequence of partial sums of a sequence \((a_n)\), given by \(s_N = \sum_{n=1}^N a_n\), converges to \(S\), then we say 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.

From this perspective, the convergence or divergence of a series reduces to that of the sequence of partial sums \(s_N\), so the tools from §Limits of Sequences apply directly. The following are two particularly important examples.

Example 2

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

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

  This formula is verified by observing that \((1-r)s_N = (1-r)(1 + r + \cdots + r^N) = 1 - r^{N+1}\). 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 either unbounded above or oscillates, and hence diverges.

• The \(p\)-series \(\sum_n 1/n^p\) converges if \(p > 1\) and diverges if \(p \leq 1\). If \(p \leq 0\), then \(1/n^p = n^{-p} \geq 1\) for all \(n\), so \(s_N \geq N\); in particular, the partial sums are unbounded above and thus diverge. Consider the case \(p > 0\). Grouping the terms of the partial sum in blocks of \(2^k\) terms (for \(k=1,\dots,m\)), excluding the first term, and looking 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 block contains \(2^k\) terms, each of which 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 therefore bounded; applying §Limits of Sequences, ⁋Proposition 7 (Monotone convergence), we see that the \(p\)-series converges. On the other hand, if \(0 < p \leq 1\), then the sum of each block is at least \(2^k\) times the smallest term \(1/(2^{k+1}-1)^p\), so

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

  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 confirm that the partial sums diverge to infinity.

The viewpoint of Definition 1, that a series reduces to the convergence problem of its sequence of partial sums, implies that the 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 is the linearity of §Limits of Sequences, ⁋Proposition 2 (Limit laws for sequences) transferred to 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

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 \(t_N \rightarrow B\), 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

In most cases, finding the exact sum of a series is too ambitious; often only convergence itself is needed. The most basic yet powerful tool for this purpose is the following proposition.

Proposition 4 (\(n\)th 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; grouping terms in blocks of \(2^k\) as before,

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

each block sums to 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 a series with positive terms, the following hold.

1. If \(0 \leq a_n \leq b_n\) for all sufficiently large \(n\) 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 if \(L < 1\) and diverges if \(L > 1\).

Proof

1: Suppose \(0 \leq a_n \leq b_n\) for \(n > N_0\), let \(t_N\) be the partial sums of \(\sum b_n\), 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 after adding the correction from the finitely many earlier terms, we see that the partial sums \((s_N)\) of \(\sum a_n\) are bounded above. Since \((s_N)\) is increasing, it converges by §Limits of Sequences, ⁋Proposition 7 (Monotone convergence).

2: If \(L < 1\), choose \(r\) with \(L < r < 1\); then 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 (\(n\)th 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 the appropriate test is important when determining convergence.

The comparison test is among the most powerful methods we examine, since it can be applied to any series provided a suitable inequality can be found; however, it carries the burden of bounding the terms directly by those of another series. In practice, it is often clear that the terms of a series are of the same order as those of some standard series, yet a clean inequality is elusive. In such cases, the following variant is convenient: it suffices to look at the limit of the ratio of terms.

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

Proof

From the definition of the limit, for all sufficiently large \(n\) we can 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 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 comparison, \(\sum b_n\) converges. The divergence case follows by contrapositive.

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 term is a fraction, one need only keep the fastest growing powers in the numerator and denominator, take this power as the standard \(b_n\), 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 \(r\) 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 comparison \(\sum a_n\) 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 (\(n\)th 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 in a sense 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. 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.