Series with terms of variable sign

Part 5 · Series · Chapter 3 · lecture notes by Fabio Furini · Chapter PDF

1. Series with terms of variable sign

Definition 1: absolutely convergent series

A series \(\sum a_k\) is absolutely convergent if the series \(\sum |a_k|\) converges.

Theorem 1

If a series \(\sum a_k\) converges absolutely, then it converges.

Proof

Without loss of generality, let \(n_0=0\) and consider the following series:

\[ \sum_{k=0}^{\infty} \big(|a_k|-a_k \big) \]

It is a series with nonnegative terms since, for every \(k\in\mathbb N\), we have:

\[ \begin{cases} |a_k|-a_k=-2a_k \ge 0 & {\rm if~~} a_k<0\\[2ex] |a_k|-a_k=0 & {\rm if~~} a_k \ge 0 \end{cases} \]

Moreover, by the triangle inequality, for every \(k\in\mathbb N\), we have:

\[ \underbrace{|a_k|-a_k}_{\ge 0}=\big||a_k|-a_k \big|\le |a_k|+|a_k|=2|a_k| \]

Hence, by the comparison test for series with nonnegative terms, we have

\[ \sum_{k=0}^\infty |a_k| {\rm~~convergent~} \Longrightarrow \sum_{k=0}^\infty \big(|a_k|-a_k \big){\rm~~convergent~~} \]

Since for every \(k\in\mathbb N\) we have \(a_k=|a_k|-\big(|a_k|-a_k\big)\) and the series \(\sum_{k=0}^\infty |a_k|\) and \(\sum_{k=0}^\infty (|a_k|-a_k)\) converge, we have:

\[ \sum_{k=0}^\infty a_k=\sum_{k=0}^\infty |a_k|-\sum_{k=0}^\infty \big(|a_k|-a_k\big) \]

that is, the series is the difference of two convergent series. Consequently, the series \(\sum_{k=0}^\infty a_k\) is convergent as well. □

• The following is an alternative proof.

Proof

Without loss of generality, let \(n_0=0\) and split the sequence \(\{s_n\}\) of partial sums into two sequences, the first containing only the positive terms and the second only the negative terms:

\[\begin{align*} s_n^+ = \sum_{\substack{k \in \{0,1,\dots,n\}:~ a_k> 0}} a_k {\rm ~~~~~and~~~~~} s_n^- = \sum_{k\in \{0,1,\dots,n\}:~ a_k< 0} -a_k {\rm ~~~~~hence~~~~~} s_n = s_n^+ - s_n^- \end{align*}\]

Consequently, it is enough to show that the sequences \(\{s_n^+\}\) and \(\{s_n^-\}\) are convergent to conclude that \(\{s_n\}\) converges, and hence that the series \(\sum a_k\) converges.

We observe that \(\{s_n^+\}\) and \(\{s_n^-\}\) are monotone non-decreasing sequences and moreover, by the triangle inequality, we have:

\[\begin{align*} s_n^+ \le \sum_{k=0}^{n} |a_k| {\rm ~~~~~and~~~~~} s_n^- \le \sum_{k=0}^{n} |a_k| \end{align*}\]

On the other hand, by hypothesis the series \(\sum a_k\) converges absolutely, i.e., \(\sum |a_k|\) converges, and hence the quantity \(\sum_{k=0}^n |a_k|\) is bounded; therefore the sequences \(\{s_n^+\}\) and \(\{s_n^-\}\) are bounded above and non-decreasing, and hence they converge, by the monotone sequence theorem. □

Example 1: absolute convergence

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} \frac{(-1)^k}{k^{\alpha}} ~~~~~~ {\rm with~~~} \alpha > 1 \]

We have:

\[ \left| \frac{(-1)^k}{k^{\alpha}} \right| = \frac{1}{k^{\alpha}}, ~\forall k \in \N, k>1 {\rm ~~~~~~and~~~~~~} \sum_{k=1}^{\infty} \frac{1}{k^{\alpha}} {\rm ~~~is~convergent~for~~~} \alpha > 1 \]

Therefore the series converges absolutely and hence it converges.

Absolute convergence implies ordinary convergence, also called simple convergence:

\[\begin{equation} \label{BBBB} \sum |a_k| {\rm ~convergent~~} ~~\Rightarrow~~ \sum a_k {\rm ~convergent~~} \end{equation}\]

but the converse is not true (we will give a counterexample later):

\[\begin{equation} \label{LLLL} \sum a_k {\rm ~convergent~~} ~~\nRightarrow~~ \sum |a_k| {\rm ~convergent~~} \end{equation}\]

Hence absolute convergence is a sufficient but not necessary condition for ordinary convergence.

1.1 Alternating series and the Leibniz test

• Among series with terms of variable sign, a particularly simple case is given by alternating series, for which the following convergence test holds.

Theorem 2: Leibniz (alternating series) test

Consider the series

\[ \sum_{k=n_0}^{\infty} (-1)^k \; a_k {\rm ~~~with~~~} a_k\ge 0, \forall k \]

If the sequence \(\{a_k\}\) is decreasing and \(a_k \rr 0\) as \(k \rr \ip\), then the series is convergent. Moreover:

\[ s_{2n} = \sum_{k=n_0}^{2n} (-1)^k \; a_k \; \downarrow \; s {\rm ~~~~~~and~~~~~~} s_{2n+1} = \sum_{k=n_0}^{2n+1} (-1)^k \; a_k \; \uparrow \; s {\rm ~~~~~as~~} n \rr \ip \]

• The partial sums with even index approximate the sum \(s\) from above and those with odd index from below.

• The Leibniz test can clearly be applied also if the terms eventually have alternating signs and the sequence \(\{a_k\}\) is eventually decreasing.

Example 2: Leibniz test

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} (-1)^k \; \frac{1}{k} \]

The sequence \(a_k = \frac{1}{k}\) is decreasing and non-negative. Moreover \(\frac{1}{k}\rr 0\) as \(k \rr \ip\), hence it satisfies the two conditions of the Leibniz test theorem. Consequently, the series is convergent.

The series \(\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\) converges and is a counterexample to \(\eqref{LLLL}\), that is, a series that converges but does not converge absolutely, since:

\[ \sum_{k=1}^{\infty} \left| \frac{(-1)^k}{k} \right| = \sum_{k=1}^{\infty} \frac{1}{k} {\rm ~~~~~and~~~~~} \sum_{k=1}^{\infty} \frac{1}{k} {\rm ~~diverges~(harmonic~series)} \]

Proof

Consider the sequence of partial sums \(\{s_n\}\) with \(n_0=0\) and the two subsequences extracted from it, \(\{ s_{2n}\}\) and \(\{s_{2n+1}\}\).

For \(\{ s_{2n}\}\) we have:

\[ s_0=a_0,~~~ s_2= s_0 - a_1 + \underbrace{a_2}_{\le a_1} \le s_0,~~~s_4= s_2 - a_3 + \underbrace{a_4}_{\le a_3} \le s_2, ~~~\dots \]

hence the sequence \(\{ s_{2n}\}\) is monotone decreasing.

For \(\{ s_{2n+1}\}\) we have:

\[ s_1=a_0-a_1,~~~ s_3= s_1 + a_2 - \underbrace{a_3}_{\le a_2} \ge s_1,~~~s_5= s_3 + a_4 - \underbrace{a_5}_{\le a_4} \ge s_3, ~~~\dots \]

hence the sequence \(\{s_{2n+1}\}\) is monotone increasing.

Moreover, we have:

\[ s_1\le s_{2n+1} = s_{2n} - a_{2n+1} \le s_{2n} \le s_0 \]

therefore \(\{s_{2n+1}\}\) is bounded above and \(\{ s_{2n}\}\) is bounded below. The two sequences are hence convergent, by the monotone sequence theorem.

The two sequences converge to the same limit, because

\[ 0 \le s_{2n} - s_{2n+1} \le a_{2n+1} \rr 0 {\rm ~~~as~~~} n \rr \ip \]

Calling \(s\) this limit, since \(\{ s_{2n}\}\) is monotone decreasing and \(\{ s_{2n+1}\}\) is monotone increasing, we have:

\[ s_{2n} = \sum_{k=n_0}^{2n} (-1)^k \; a_k \; \downarrow \; s {\rm ~~~~~~and~~~~~~} s_{2n+1} = \sum_{k=n_0}^{2n+1} (-1)^k \; a_k \; \uparrow \; s {\rm ~~~~~as~~} n \rr \ip \]

Hence the series is convergent since \(s_{2n+1} \rr s~\) and \(s_{2n} \rr s~\) as \(n \rr \ip\), and consequently we have:

\[ s_n \rr s {\rm ~~~as~~~} n \rr \ip \]

□

Corollary 1: of the Leibniz test theorem

Given a series satisfying the hypotheses of the Leibniz test theorem, we have:

\[ \underbrace{ |s-s_{m}|}_{=\left| \sum_{k=m+1}^{\infty} (-1)^k \; a_k \right|} \le a_{m+1}, ~~~\forall m \in \N \]

• For every \(m\), the error made by approximating \(s\) with \(s_m\) is, in absolute value, bounded above by the value of the first omitted term. In other words, the tail of the series tends to a value less than or equal to \(a_{m+1}\).

Proof

Since

\[ s_{2n+1} \; \uparrow \; s {\rm ~~~~~~and~~~~~~} s_{2n} \; \downarrow \; s {\rm ~~~as~~} n \rr \ip \]

we therefore have, for every \(n \in \N\):

\[ s_{2n-1} \le s \le s_{2n} {\rm ~~~~~~and~~~~~~} s_{2n+1} \le s \le s_{2n} \]

from which we deduce

\[ 0 \le s - s_{2n-1} \le s_{2n} - s_{2n-1} = a_{2n} {\rm ~~~~~~and~~~~~~} 0 \le s_{2n} - s \le s_{2n} - s_{2n+1} = a_{2n+1} \]

Therefore, for every \(m\), whether even or odd, we have:

\[ |s-s_{m}|=\left| \sum_{k=m+1}^{\infty} (-1)^k \; a_k \right| \le a_{m+1} \]

□

Example 3: Leibniz test

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} (-1)^k \; \frac{k-1 }{k^2+k} \]

We have:

\[ a_k=\frac{k-1 }{k^2+k} = \frac{k-1 }{k\;(k+1)} \ge 0, \forall k \in \N, k \ge 1 \qquad{\rm and}\qquad \frac{k-1 }{k\;(k+1)} \sim \frac{1}{k} \rr 0 {\rm ~~as~~} k \rr \ip \]

hence the series does not converge absolutely. However, the series is decreasing for \(k \ge 2\) since:

\[ \underbrace{\frac{k}{(k+1)(k+2)}}_{= a_{k+1}} \le \underbrace{\frac{k-1 }{k\;(k+1)}}_{= a_k} \Longleftrightarrow k^2 \le k^2 + k -2 \Longleftrightarrow k \ge 2 \]

hence the series converges by the Leibniz test, since it is eventually decreasing.

Example 4: Leibniz test

Let us determine the behavior of the series:

\[ \sum_{k=2}^{\infty} (-1)^k \; \frac{\log k}{k} \]

We have:

\[ a_k= \frac{\log k}{k} \ge 0, \forall k \ge 2 {\rm ~~~~~and~~~~~} \frac{\log k}{k} \rr 0 {\rm ~~as~~} k \rr \ip \]

moreover

\[ \frac{\log k}{k} > \frac{1}{k} {\rm ~~~for~~} k \ge 3 {\rm ~~~~~and~~~~~} \sum_{k=1}^{\infty} \frac{1}{k} {\rm ~~~~diverges} \]

hence the series does not converge absolutely. Proving algebraically that the sequence \(\{a_k\}\) is decreasing is complicated; instead, we pass from the discrete to the continuous setting. We have, for \(x \in \R\) and \(x \ge 2\):

\[ f(x) = \frac{\log x}{x} {\rm ~~~~and~~~~} f'(x) = \frac{1-\log x}{x^2} \le 0 {\rm ~~~for~~~} x \ge e \]

It follows that \(f\) is decreasing for \(x \ge e\); consequently the sequence \(a_k = f ( k)\) is decreasing for \(k \ge 3\) (the first integer \(> e\)). Hence the series converges by the Leibniz test.

\[ {\rm if~~} \sum a_k {\rm ~converges~~~and~~~~} \sum b_k {\rm ~converges~~~~~then~~} \sum (a_k+b_k) {\rm ~converges} \]

\[ {\rm if~~} \sum a_k {\rm ~converges~~~and~~~~} \sum b_k {\rm ~diverges~~~~~then~~} \sum (a_k+b_k) {\rm ~diverges} \]

This can be checked by viewing the series as the limit of the sequence of partial sums and applying the theorem on the limit of a sum.

Example 5: Leibniz test

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} \frac{k+1 +(-1)^k \; k^2}{k^3} \]

We split it into the sum of two series:

\[ \sum_{k=1}^{\infty} \frac{k+1 +(-1)^k \; k^2}{k^3} = \sum_{k=1}^{\infty} \frac{k+1 }{k^3} + \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \]

For the first series we have:

\[ \frac{k+1 }{k^3} \ge 0, \forall k \ge 1 \qquad{\rm and}\qquad \frac{k+1 }{k^3}\rr 0 {\rm ~~as~~} k \rr \ip \]

hence it converges by the limit comparison test, since:

\[ \frac{k+1 }{k^3} \sim \frac{1}{k^2} {\rm ~~~~~and~~~~~} \sum_{k=1}^{\infty} \frac{1}{k^2} {\rm ~~~~converges} \]

The second series converges by the Leibniz test; hence the original series converges.

Example 6: Leibniz test

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} (-1)^k \left( \frac{\sqrt{k}+(-1)^k}{k}\right) \]

We split it into the sum of two series:

\[ \sum_{k=1}^{\infty} (-1)^k \left( \frac{\sqrt{k}+(-1)^k}{k}\right) = \sum_{k=1}^{\infty} \frac{(-1)^k}{\sqrt{k}} + \sum_{k=1}^{\infty} \frac{1}{k} \]

The first series converges by the Leibniz test; the second diverges (harmonic series); hence the original series diverges.

Example 7: Leibniz test

Let us determine the behavior of the series:

\[ \sum_{k=1}^{\infty} \frac{(-1)^{k+1} }{k} \]

We have:

\[ \frac{(-1)^{k+1} }{k} = -\frac{(-1)^{k} }{k}, ~\forall k\in \N, k>1 \]

moreover, the sequence \(a_k = \frac{1}{k}\) is decreasing and \(\frac{1}{k}\rr 0\) as \(k \rr \ip\), hence it satisfies the two conditions of the theorem. Consequently, the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) converges.

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