Section 2.5 Series
Note: 2 lectures
A fundamental object in mathematics is that of a series. In fact, when the foundations of analysis were being developed, the motivation was to understand series. Understanding series is important in applications of analysis. For example, solutions to differential equations are often given as series, and differential equations are the basis for understanding almost all of modern science.
Subsection 2.5.1 Definition
Definition 2.5.1.
Given a sequence \(\{ x_n \}\text{,}\) we write the formal object
and call it a series. A series converges if the sequence \(\{ s_k \}\) defined by
converges. The numbers \(s_k\) are called partial sums. If \(x := \lim\, s_k\text{,}\) we write
In this case, we cheat a little and treat \(\sum_{n=1}^\infty x_n\) as a number.
If the sequence \(\{ s_k \}\) diverges, we say the series is divergent. In this case, \(\sum x_n\) is simply a formal object and not a number.
In other words, for a convergent series, we have
We only have this equality if the limit on the right actually exists. If the series does not converge, the righthand side does not make sense (the limit does not exist). Therefore, be careful as \(\sum x_n\) means two different things (a notation for the series itself or the limit of the partial sums), and you must use context to distinguish.
Remark 2.5.2.
It is sometimes convenient to start the series at an index different from 1. For instance, we can write
The lefthand side is more convenient to write.
Remark 2.5.3.
It is common to write the series \(\sum x_n\) as
with the understanding that the ellipsis indicates a series and not a simple sum. We do not use this notation as it is the sort of informal notation that leads to mistakes in proofs.
Example 2.5.4.
The series
converges and the limit is 1. That is,
Proof: First we prove the following equality
The equality is immediate when \(k=1\text{.}\) The proof for general \(k\) follows by induction, which we leave to the reader. See Figure 2.7 for an illustration.
Let \(s_k\) be the partial sum. We write
The sequence \(\bigl\{ \frac{1}{2^k} \bigr\}\text{,}\) and therefore \(\bigl\{ \abs{1s_k} \bigr\}\text{,}\) converges to zero. So, \(\{ s_k \}\) converges to 1.
Proposition 2.5.5.
Suppose \(1 < r < 1\text{.}\) Then the geometric series \(\sum_{n=0}^\infty r^n\) converges, and
Details of the proof are left as an exercise. The proof consists of showing
and then taking the limit as \(k\) goes to \(\infty\text{.}\) Geometric series is one of the most important series, and in fact it is one of the few series for which we can so explicitly find the limit.
As for sequences we can talk about a tail of a series.
Proposition 2.5.6.
Let \(\sum x_n\) be a series. Let \(M \in \N\text{.}\) Then
Proof.
We look at partial sums of the two series (for \(k \geq M\))
Note that \(\sum_{n=1}^{M1} x_n\) is a fixed number. Use Proposition 2.2.5 to finish the proof.
Subsection 2.5.2 Cauchy series
Definition 2.5.7.
A series \(\sum x_n\) is said to be Cauchy or a Cauchy series if the sequence of partial sums \(\{ s_n \}\) is a Cauchy sequence.
A sequence of real numbers converges if and only if it is Cauchy. Therefore, a series is convergent if and only if it is Cauchy. The series \(\sum x_n\) is Cauchy if for every \(\epsilon > 0\text{,}\) there exists an \(M \in \N\text{,}\) such that for every \(n \geq M\) and \(k \geq M\text{,}\) we have
Without loss of generality we assume \(n < k\text{.}\) Then we write
We have proved the following simple proposition.
Proposition 2.5.8.
The series \(\sum x_n\) is Cauchy if for every \(\epsilon > 0\text{,}\) there exists an \(M \in \N\) such that for every \(n \geq M\) and every \(k > n\text{,}\) we have
Subsection 2.5.3 Basic properties
Proposition 2.5.9.
Let \(\sum x_n\) be a convergent series. Then the sequence \(\{ x_n \}\) is convergent and
Proof.
Let \(\epsilon > 0\) be given. As \(\sum x_n\) is convergent, it is Cauchy. Thus we find an \(M\) such that for every \(n \geq M\text{,}\) we have
Hence for every \(n \geq M+1\text{,}\) we have \(\abs{x_{n}} < \epsilon\text{.}\)
Example 2.5.10.
If \(r \geq 1\) or \(r \leq 1\text{,}\) then the geometric series \(\sum_{n=0}^\infty r^n\) diverges.
Proof: \(\abs{r^n} = \abs{r}^n \geq 1^n = 1\text{.}\) So the terms do not go to zero and the series cannot converge.
So if a series converges, the terms of the series go to zero. The implication, however, goes only one way. Let us give an example.
Example 2.5.11.
The series \(\sum \frac{1}{n}\) diverges (despite the fact that \(\lim \frac{1}{n} = 0\)). This is the famous harmonic series^{ 1 }.
Proof: We will show that the sequence of partial sums is unbounded, and hence cannot converge. Write the partial sums \(s_n\) for \(n = 2^k\) as:
Notice \(\nicefrac{1}{3} + \nicefrac{1}{4} \geq \nicefrac{1}{4} + \nicefrac{1}{4} = \nicefrac{1}{2}\) and \(\nicefrac{1}{5} + \nicefrac{1}{6} + \nicefrac{1}{7} + \nicefrac{1}{8} \geq \nicefrac{1}{8} + \nicefrac{1}{8} + \nicefrac{1}{8} + \nicefrac{1}{8} = \nicefrac{1}{2}\text{.}\) More generally
Therefore,
As \(\{ \frac{k}{2} \}\) is unbounded by the Archimedean property, that means that \(\{ s_{2^k} \}\) is unbounded, and therefore \(\{ s_n \}\) is unbounded. Hence \(\{ s_n \}\) diverges, and consequently \(\sum \frac{1}{n}\) diverges.
Convergent series are linear. That is, we can multiply them by constants and add them and these operations are done term by term.
Proposition 2.5.12. Linearity of series.
Let \(\alpha \in \R\) and \(\sum x_n\) and \(\sum y_n\) be convergent series. Then

\(\sum \alpha x_n\) is a convergent series and
\begin{equation*} \sum_{n=1}^\infty \alpha x_n = \alpha \sum_{n=1}^\infty x_n . \end{equation*} 
\(\sum ( x_n + y_n )\) is a convergent series and
\begin{equation*} \sum_{n=1}^\infty ( x_n + y_n ) = \left( \sum_{n=1}^\infty x_n \right) + \left( \sum_{n=1}^\infty y_n \right) . \end{equation*}
Proof.
For the first item, we simply write the \(k\)th partial sum
We look at the righthand side and note that the constant multiple of a convergent sequence is convergent. Hence, we take the limit of both sides to obtain the result.
For the second item we also look at the \(k\)th partial sum
We look at the righthand side and note that the sum of convergent sequences is convergent. Hence, we take the limit of both sides to obtain the proposition.
An example of a useful application of the first item is the following formula. If \(\abs{r} < 1\) and \(j \in \N\text{,}\) then
The formula follows by using the geometric series and multiplying by \(r^j\text{:}\)
Multiplying series is not as simple as adding, see the next section. It is not true, of course, that we multiply term by term. That strategy does not work even for finite sums: \((a+b)(c+d) \not= ac+bd\text{.}\)
Subsection 2.5.4 Absolute convergence
As monotone sequences are easier to work with than arbitrary sequences, it is usually easier to work with series \(\sum x_n\text{,}\) where \(x_n \geq 0\) for all \(n\text{.}\) The sequence of partial sums is then monotone increasing and converges if it is bounded above. Let us formalize this statement as a proposition.
Proposition 2.5.13.
If \(x_n \geq 0\) for all \(n\text{,}\) then \(\sum x_n\) converges if and only if the sequence of partial sums is bounded above.
As the limit of a monotone increasing sequence is the supremum, then when \(x_n \geq 0\) for all \(n\text{,}\) we have the inequality
If we allow infinite limits, the inequality still holds even when the series diverges to infinity, although in that case it is not terribly useful.
We will see that the following common criterion for convergence of series has big implications for how the series can be manipulated.
Definition 2.5.14.
A series \(\sum x_n\) converges absolutely if the series \(\sum \abs{x_n}\) converges. If a series converges, but does not converge absolutely, we say it is conditionally convergent.
Proposition 2.5.15.
If the series \(\sum x_n\) converges absolutely, then it converges.
Proof.
A series is convergent if and only if it is Cauchy. Hence suppose \(\sum \abs{x_n}\) is Cauchy. That is, for every \(\epsilon > 0\text{,}\) there exists an \(M\) such that for all \(k \geq M\) and all \(n > k\text{,}\) we have
We apply the triangle inequality for a finite sum to obtain
Hence \(\sum x_n\) is Cauchy, and therefore it converges.
If \(\sum x_n\) converges absolutely, the limits of \(\sum x_n\) and \(\sum \abs{x_n}\) are generally different. Computing one does not help us compute the other. However, the computation above leads to a useful inequality for absolutely convergent series, a series version of the triangle inequality, a proof of which we leave as an exercise:
Absolutely convergent series have many wonderful properties. For example, absolutely convergent series can be rearranged arbitrarily, or we can multiply such series together easily. Conditionally convergent series on the other hand often do not behave as one would expect. See the next section.
We leave as an exercise to show that
converges, although the reader should finish this section before trying. On the other hand, we proved
diverges. Therefore, \(\sum \frac{{(1)}^n}{n}\) is a conditionally convergent series.
Subsection 2.5.5 Comparison test and the \(p\)series
We noted above that for a series to converge the terms not only have to go to zero, but they have to go to zero “fast enough.” If we know about convergence of a certain series, we can use the following comparison test to see if the terms of another series go to zero “fast enough.”
Proposition 2.5.16. Comparison test.
Let \(\sum x_n\) and \(\sum y_n\) be series such that \(0 \leq x_n \leq y_n\) for all \(n \in \N\text{.}\)
If \(\sum y_n\) converges, then so does \(\sum x_n\text{.}\)
If \(\sum x_n\) diverges, then so does \(\sum y_n\text{.}\)
Proof.
As the terms of the series are all nonnegative, the sequences of partial sums are both monotone increasing. Since \(x_n \leq y_n\) for all \(n\text{,}\) the partial sums satisfy for all \(k\)
If the series \(\sum y_n\) converges, the partial sums for the series are bounded. Therefore, the righthand side of (2.1) is bounded for all \(k\text{;}\) there exists some \(B \in \R\) such that \(\sum_{n=1}^k y_n \leq B\) for all \(k\text{,}\) and so
Hence the partial sums for \(\sum x_n\) are also bounded. Since the partial sums are a monotone increasing sequence they are convergent. The first item is thus proved.
On the other hand if \(\sum x_n\) diverges, the sequence of partial sums must be unbounded since it is monotone increasing. That is, the partial sums for \(\sum x_n\) are eventually bigger than any real number. Putting this together with (2.1) we see that for every \(B \in \R\text{,}\) there is a \(k\) such that
Hence the partial sums for \(\sum y_n\) are also unbounded, and \(\sum y_n\) also diverges.
A useful series to use with the comparison test is the \(p\)series^{ 3 }.
Proposition 2.5.17. \(p\)series or the \(p\)test.
For \(p \in \R\text{,}\) the series
converges if and only if \(p > 1\text{.}\)
Proof.
First suppose \(p \leq 1\text{.}\) As \(n \geq 1\text{,}\) we have \(\frac{1}{n^p} \geq \frac{1}{n}\text{.}\) Since \(\sum \frac{1}{n}\) diverges, \(\sum \frac{1}{n^p}\) must diverge for all \(p \leq 1\) by the comparison test.
Now suppose \(p > 1\text{.}\) We proceed as we did for the harmonic series, but instead of showing that the sequence of partial sums is unbounded, we show that it is bounded. The terms of the series are positive, so the sequence of partial sums is monotone increasing and converges if it is bounded above. Let \(s_n\) denote the \(n\)th partial sum.
Instead of estimating from below, we estimate from above. As \(p\) is positive, then \(2^p < 3^p\text{,}\) and hence \(\frac{1}{2^p} + \frac{1}{3^p} < \frac{1}{2^p} + \frac{1}{2^p}\text{.}\) Similarly, \(\frac{1}{4^p} + \frac{1}{5^p} + \frac{1}{6^p} + \frac{1}{7^p} < \frac{1}{4^p} + \frac{1}{4^p} + \frac{1}{4^p} + \frac{1}{4^p}\text{.}\) Therefore, for all \(k \geq 2\text{,}\)
As \(p > 1\text{,}\) then \(\frac{1}{2^{p1}} < 1\text{.}\) Proposition 2.5.5 says that
converges. Thus,
For every \(n\) there is a \(k \geq 2\) such that \(n \leq 2^k1\text{,}\) and as \(\{ s_n \}\) is a monotone sequence, \(s_n \leq s_{2^k1}\text{.}\) So for all \(n\text{,}\)
Thus the sequence of partial sums is bounded, and the series converges.
Neither the \(p\)series test nor the comparison test tell us what the sum converges to. They only tell us that a limit of the partial sums exists. For instance, while we know that \(\sum \nicefrac{1}{n^2}\) converges, it is far harder to find^{ 4 } that the limit is \(\nicefrac{\pi^2}{6}\text{.}\) If we treat \(\sum \nicefrac{1}{n^p}\) as a function of \(p\text{,}\) we get the socalled Riemann \(\zeta\) function. Understanding the behavior of this function contains one of the most famous unsolved problems in mathematics today and has applications in seemingly unrelated areas such as modern cryptography.
Example 2.5.18.
The series \(\sum \frac{1}{n^2+1}\) converges.
Proof: First, \(\frac{1}{n^2+1} < \frac{1}{n^2}\) for all \(n \in \N\text{.}\) The series \(\sum \frac{1}{n^2}\) converges by the \(p\)series test. Therefore, by the comparison test, \(\sum \frac{1}{n^2+1}\) converges.
Subsection 2.5.6 Ratio test
Suppose \(r > 0\text{.}\) The ratio of two subsequent terms in the geometric series \(\sum r^n\) is \(\frac{r^{n+1}}{r^n} = r\text{,}\) and the series converges whenever \(r < 1\text{.}\) Just as for sequences, this fact can be generalized to more arbitrary series as long as we have such a ratio “in the limit.” We then compare the tail of a series to the geometric series.
Proposition 2.5.19. Ratio test.
Let \(\sum x_n\) be a series, \(x_n \not= 0\) for all \(n\text{,}\) and such that
If \(L < 1\text{,}\) then \(\sum x_n\) converges absolutely.
If \(L > 1\text{,}\) then \(\sum x_n\) diverges.
Although the test as stated is often sufficient, it can be strengthened a bit, see Exercise 2.5.6.
Proof.
If \(L > 1\text{,}\) then Lemma 2.2.12 says that the sequence \(\{ x_n \}\) diverges. Since it is a necessary condition for the convergence of series that the terms go to zero, we know that \(\sum x_n\) must diverge.
Thus suppose \(L < 1\text{.}\) We will argue that \(\sum \abs{x_n}\) must converge. The proof is similar to that of Lemma 2.2.12. Of course \(L \geq 0\text{.}\) Pick \(r\) such that \(L < r < 1\text{.}\) As \(rL > 0\text{,}\) there exists an \(M \in \N\) such that for all \(n \geq M\text{,}\)
Therefore,
For \(n > M\) (that is for \(n \geq M+1\)), write
For \(k > M\text{,}\) write the partial sum as
As \(0 < r < 1\text{,}\) the geometric series \(\sum_{n=0}^{\infty} r^n\) converges, so \(\sum_{n=M+1}^{\infty} r^n\) converges as well. We take the limit as \(k\) goes to infinity on the righthand side above to obtain
The righthand side is a number that does not depend on \(k\text{.}\) Hence the sequence of partial sums of \(\sum \abs{x_n}\) is bounded and \(\sum \abs{x_n}\) is convergent. Thus \(\sum x_n\) is absolutely convergent.
Example 2.5.20.
The series
converges absolutely.
Proof: We write
Therefore, the series converges absolutely by the ratio test.
Subsection 2.5.7 Exercises
Exercise 2.5.1.
Suppose the \(k\)th partial sum of \(\displaystyle \sum_{n=1}^\infty x_n\) is \(s_k = \frac{k}{k+1}\text{.}\) Find the series, that is find \(x_n\text{,}\) prove that the series converges, and then find the limit.
Exercise 2.5.2.
Prove Proposition 2.5.5, that is for \(1 < r < 1\) prove
Hint: See Example 0.3.8.
Exercise 2.5.3.
Decide the convergence or divergence of the following series.
a) \(\displaystyle \sum_{n=1}^\infty \frac{3}{9n+1}\) b) \(\displaystyle \sum_{n=1}^\infty \frac{1}{2n1}\) c) \(\displaystyle \sum_{n=1}^\infty \frac{{(1)}^n}{n^2}\) d) \(\displaystyle \sum_{n=1}^\infty \frac{1}{n(n+1)}\) e) \(\displaystyle \sum_{n=1}^\infty n e^{n^2}\)
Exercise 2.5.4.
Prove that if \(\displaystyle \sum_{n=1}^\infty x_n\) converges, then \(\displaystyle \sum_{n=1}^\infty ( x_{2n} + x_{2n+1} )\) also converges.
Find an explicit example where the converse does not hold.
Exercise 2.5.5.
For \(j=1,2,\ldots,n\text{,}\) let \(\{ x_{j,k} \}_{k=1}^\infty\) denote \(n\) sequences. Suppose that for each \(j \in \N\text{,}\)
is convergent. Prove
Exercise 2.5.6.
Prove the following stronger version of the ratio test: Let \(\sum x_n\) be a series.
If there is an \(N\) and a \(\rho < 1\) such that \(\frac{\abs{x_{n+1}}}{\abs{x_n}} < \rho\) for all \(n \geq N\text{,}\) then the series converges absolutely. (Remark: Equivalently the condition can be stated as \(\limsup_{n\to\infty} \frac{\abs{x_{n+1}}}{\abs{x_n}} < 1\text{.}\))
If there is an \(N\) such that \(\frac{\abs{x_{n+1}}}{\abs{x_n}} \geq 1\) for all \(n \geq N\text{,}\) then the series diverges.
Exercise 2.5.7.
(Challenging) Suppose \(\{ x_n \}\) is a decreasing sequence and \(\sum x_n\) converges. Prove \(\displaystyle \lim_{n\to\infty} n x_n = 0\text{.}\)
Exercise 2.5.8.
Show that \(\displaystyle \sum_{n=1}^\infty \frac{{(1)}^n}{n}\) converges. Hint: Consider the sum of two subsequent entries.
Exercise 2.5.9.
Prove that if \(\sum x_n\) and \(\sum y_n\) converge absolutely, then \(\sum x_ny_n\) converges absolutely.
Find an explicit example where the converse does not hold.
Find an explicit example where all three series are absolutely convergent, are not just finite sums, and \((\sum x_n)(\sum y_n) \not= \sum x_ny_n\text{.}\) That is, show that series are not multiplied termbyterm.
Exercise 2.5.10.
Prove the triangle inequality for series: If \(\sum x_n\) converges absolutely, then
Exercise 2.5.11.
Prove the limit comparison test. That is, prove that if \(a_n > 0\) and \(b_n > 0\) for all \(n\text{,}\) and
then either \(\sum a_n\) and \(\sum b_n\) both converge or both diverge.
Exercise 2.5.12.
Let \(x_n := \sum_{j=1}^n \nicefrac{1}{j}\text{.}\) Show that for every \(k\text{,}\) we get \(\displaystyle \lim_{n\to\infty} \abs{x_{n+k}x_n} = 0\text{,}\) yet \(\{ x_n \}\) is not Cauchy.
Exercise 2.5.13.
Let \(s_k\) be the \(k\)th partial sum of \(\sum x_n\text{.}\)
Suppose that there exists an \(m \in \N\) such that \(\displaystyle \lim_{k\to\infty} s_{mk}\) exists and \(\lim\, x_n = 0\text{.}\) Show that \(\sum x_n\) converges.
Find an example where \(\displaystyle \lim_{k\to\infty} s_{2k}\) exists and \(\lim\, x_n \not= 0\) (and therefore \(\sum x_n\) diverges).
(Challenging) Find an example where \(\lim\, x_n = 0\text{,}\) and there exists a subsequence \(\{ s_{k_j} \}\) such that \(\displaystyle \lim_{j\to\infty} s_{k_j}\) exists, but \(\sum x_n\) still diverges.
Exercise 2.5.14.
Suppose \(\sum x_n\) converges and \(x_n \geq 0\) for all \(n\text{.}\) Prove that \(\sum x_n^2\) converges.
Exercise 2.5.15.
(Challenging) Suppose \(\{ x_n\}\) is a decreasing sequence of positive numbers. The proof of convergence/divergence for the \(p\)series generalizes. Prove the socalled Cauchy condensation principle:
Exercise 2.5.16.
Use the Cauchy condensation principle (see Exercise 2.5.15) to decide the convergence of
a) \(\displaystyle \sum \frac{\ln n}{n^2}\) b) \(\displaystyle \sum \frac{1}{n \ln n}\) c) \(\displaystyle \sum \frac{1}{n {(\ln n)}^2}\) d) \(\displaystyle \sum \frac{1}{n (\ln n ){(\ln \ln n)}^2}\)
For the series to be welldefined you need to start some of the series at \(n=2\text{.}\) Note that only the tails of some of these series satisfy the hypotheses of the principle; you should argue why that is sufficient.
Hint: Feel free to use the identity \(\ln (2^n) = n \ln 2\text{.}\)
Exercise 2.5.17.
(Challenging) Prove Abel's theorem:
Theorem. Suppose \(\sum x_n\) is a series whose partial sums are a bounded sequence, \(\{ \lambda_n \}\) is a sequence with \(\lim \lambda_n = 0\text{,}\) and \(\sum \abs{ \lambda_{n+1}  \lambda_n }\) is convergent. Then \(\sum \lambda_n x_n\) is convergent.
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