# 2. Series

## 2. Series

Table of Content

### Convergence

#### Convergence

If the sequence of partial sums \((s_n)\) has a limit \(s\in \mathbb{R}\), then
the **series** of the sequence \((a_k)\) **converges**
and its sum is \(s\). This is denoted by
\[
a_1+a_2+\dots =\sum_{k=1}^{\infty} a_k = \lim_{n\to\infty}\underbrace{\sum_{k=1}^{n} a_k}_{=s_{n}} = s.
\]

#### Indexing

The partial sums should be indexed in the same way as the sequence \((a_k)\); e.g. the partial sums of a sequence \((a_k)_{k=0}^{\infty}\) are \(s_0= a_0, s_1=a_0+a_1\) etc.

The indexing of a series can be shifted without altering the series: \[\sum_{k=1}^{\infty} a_k =\sum_{k=0}^{\infty} a_{k+1} = \sum_{k=2}^{\infty} a_{k-1}.\]

In a concrete way: \[\sum_{k=1}^{\infty} \frac{1}{k^2}=1+\frac{1}{4}+\frac{1}{9}+\dots= \sum_{k=0}^{\infty} \frac{1}{(k+1)^2}\]

**Interactivity.**

Compute partial sums of the series \(\displaystyle\sum_{k=0}^{\infty}a_{k}\)

\(k\)th-element of the series: , start summation at

#### Divergence of a series

A series that does ot converge is **divergent**. This can
happen in three different ways:

- the partial sums tend to infinity
- the partial sums tend to minus infinity
- the sequence of partial sums oscillates so that there is no limit.

In the case of a divergent series the symbol \(\displaystyle\sum_{k=1}^{\infty} a_k\) does not really mean anything (it isn't a number). We can then interpret it as the sequence of partial sums, which is always well-defined.

### Basic results

#### Geometric series

A geometric series \[\sum_{k=0}^{\infty} aq^k\] converges if \(|q|<1\) (or \(a=0\)), and then its sum is \(\frac{a}{1-q}\). If \(|q|\ge 1\), then the series diverges.

**Proof.** The partial sums satisfy
\[\sum_{k=0}^{n} aq^k =\frac{a(1-q^{n+1})}{1-q},\]
from which the claim follows.

\(\square\)

More generally \[\sum_{k=i}^{\infty} aq^k = \frac{aq^i}{1-q} = \frac{\text{1st term of the series}}{1-q},\text{ for } |q|<1.\]

##### Example 1.

Calculate the sum of the series \[\sum_{k=1}^{\infty}\frac{3}{4^{k+1}}.\]

**Solution.** Since
\[\frac{3}{4^{k+1}} = \frac{3}{4}\cdot \left( \frac{1}{4}\right)^k,\]
this is a geometric series. The sum is
\[\frac{3}{4}\cdot \frac{1/4}{1-1/4} = \frac{1}{4}.\]

#### Rules of summation

Properties of convergent series:
- \(\displaystyle{\sum_{k=1}^{\infty} (a_k+b_k) = \sum_{k=1}^{\infty} a_k + \sum_{k=1}^{\infty} b_k}\)
- \(\displaystyle{\sum_{k=1}^{\infty} (c\, a_k) = c\sum_{k=1}^{\infty} a_k}\), where \(c\in \mathbb{R}\) is a constant

**Proof.** These follow from the corresponding properties for limits of a sequence.

\(\square\)

Note: Compared to limits, there is no similar product-rule for series, because even for sums of two elements we have \[(a_1+a_2)(b_1+b_2) \neq a_1b_1 +a_2b_2.\] The correct generalization is the Cauchy product of two series, where also the cross terms are taken into account.

##### Theorem 1.

If the series \(\displaystyle{\sum_{k=1}^{\infty} a_k}\) converges, then \[\displaystyle{\lim_{k\to \infty} a_k =0}.\]

Conversely: If \[\displaystyle{\lim_{k\to \infty} a_k \neq 0},\] then the series \(\displaystyle{\sum_{k=1}^{\infty} a_k}\) diverges.

If the sum of the series is \(s\), then \(a_k=s_k-s_{k-1}\to s-s=0\).

\(\square\)

**Note:** The property \(\lim_{k\to \infty} a_k = 0\)
cannot be used to justify the convergence of a series; cf. the following examples.
This is one of the most common elementary mistakes many people do when
studying series!

##### Example

Explore the convergence of the series \[\sum_{k=1}^{\infty} \frac{k}{k+1} = \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\dots\]

**Solution.** The limit of the general term of the series is
\[\lim_{k\to\infty}\frac{k}{k+1} = 1.\]
As this is different from zero, the series diverges.

#### Harmonic series

The harmonic series \[\sum_{k=1}^{\infty} \frac{1}{k} = 1+\frac{1}{2}+\frac{1}{3}+\dots\] diverges, although the limit of the general term \(a_k=1/k\) equals zero.

This is a classical result first proven in the 14th century by Nicole Oresme after which a number of proofs using different approaches have been published. Here we present two different approaches for comparison.

**i) An elementary proof by contradiction.** Suppose, for the sake of contradiction, that the harmonic series converges i.e. there exists \(s\in\mathbb{R}\)
such that \(s = \sum_{k=1}^{\infty}1/k\). In this case
\[
s = \left(\color{#4334eb}{1} + \color{#eb7134}{\frac{1}{2}}\right) + \left(\color{#4334eb}{\frac{1}{3}} +
\color{#eb7134}{\frac{1}{4}}\right) + \left(\color{#4334eb}{\frac{1}{5}} + \color{#eb7134}{\frac{1}{6}}\right) + \dots
= \sum_{k=1}^{\infty}\left(\color{#4334eb}{\frac{1}{2k-1}} + \color{#eb7134}{\frac{1}{2k}}\right).
\]
Now, by direct comparison we get
\[
\color{#4334eb}{\frac{1}{2k-1}} > \color{#eb7134}{\frac{1}{2k}} > 0, \text{ for all }k\ge 1~\Rightarrow~\sum_{k=1}^{\infty}\color{#4334eb}{\frac{1}{2k-1}} > \sum_{k=1}^{\infty}\color{#eb7134}{\frac{1}{2k}} = \frac{s}{2}
\]
hence following from the Properties of summation it follows that
\[
s = \sum_{k=1}^{\infty}\color{#4334eb}{\frac{1}{2k-1}} + \sum_{k=1}^{\infty}\color{#eb7134}{\frac{1}{2k}} = \sum_{k=1}^{\infty}\color{#4334eb}{\frac{1}{2k-1}} + \frac{1}{2}\underbrace{\sum_{k=1}^{\infty}\frac{1}{k}}_{=s}.
\]
\[
= \sum_{k=1}^{\infty}\color{#4334eb}{\frac{1}{2k-1}} + \frac{s}{2} > \sum_{k=1}^{\infty}\color{#eb7134}{\frac{1}{2k}} + \frac{s}{2} = \frac{s}{2} + \frac{s}{2} = s.
\]
But this implies that \(s>s\), a contradiction. Therefore, the initial assumption that the harmonic series converges must be false and thus the series diverges.

\(\square\)

**ii) Proof using integral:** Below a histogram with heights \(1/k\) lies the graph of
the function \(f(x)=1/(x+1)\), so comparing areas we have
\[\sum_{k=1}^{n} \frac{1}{k} \ge \int_0^n\frac{dx}{x+1} =\ln(n+1)\to\infty, \]
as \(n\to\infty\).

\(\square\)

#### Positive series

Summing a series is often difficult or even impossible in closed form, sometimes only a numerical approximation can be calculated. The first goal then is to find out whether a series is convergent or divergent.

A series \(\displaystyle{\sum_{k=1}^{\infty} p_k}\) is **positive**,
if \(p_k > 0\) for all \(k\).

Convergence of positive series is quite straightforward:

##### Theorem 2.

A positive series converges if and only if the sequence of partial sums is bounded from above.

*Why?* Because the partial sums form an increasing sequence.

##### Example

Show that the partial sums of a *superharmonic* series
\[\sum_{k=1}^{\infty}\frac{1}{k^2}\]
satisfy \(s_n<2\) for all \(n\), so the series converges.

**Solution.** This is based on the formula
\[\frac{1}{k^2} < \frac{1}{k(k-1)} = \frac{1}{k-1}-\frac{1}{k},\]
for \(k\ge 2\), as it implies that
\[\sum_{k=1}^n\frac{1}{k^2} < 1+ \sum_{k=2}^n\frac{1}{k(k-1)} =2-\frac{1}{n}< 2\]
for all \(n\ge 2\).

This can also be proven with integrals.

Leonhard Euler found out in 1735 that the sum is actually \(\pi^2/6\). His proof was based on comparison of the series and product expansion of the sine function.

### Absolute convergence

##### Definition

A series \(\displaystyle{\sum_{k=1}^{\infty} a_k}\) converges **absolutely** if
the positive series \(\sum_{k=1}^{\infty} |a_k|\) converges.

##### Theorem 3.

An absolutely convergent series converges (in the usual sense) and \[\left| \sum_{k=1}^{\infty} a_k \right| \le \sum_{k=1}^{\infty} |a_k|.\]

This is a special case of the *Comparison principle*, see later.

Suppose that \(\sum_k |a_k|\) converges. We study separately the positive and negative
parts of \(\sum_k a_k\):
Let \[b_k=\max (a_k,0)\ge 0 \text{ and } c_k=-\min (a_k,0)\ge 0.\]
Since \(b_k,c_k\le |a_k|\), the positive series \(\sum b_k\) and \(\sum c_k\) converge by Theorem 2.
Also, \(a_k=b_k-c_k\), so \(\sum a_k\) converges as a difference of two convergent series.

\(\square\)

##### Example

Study the convergence of the alternating (= the signs alternate) series \[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}=1-\frac{1}{4}+\frac{1}{9}-\dots\]

**Solution.** Since
\[\displaystyle{\left| \frac{(-1)^{k+1}}{k^2}\right| =\frac{1}{k^2}}\]
and the superharmonic series
\[\sum_{k=1}^{\infty}\frac{1}{k^2}\]
converges, then the original series is absolutely convergent. Therefore
it also converges in the usual sense.

#### Alternating harmonic series

The usual convergence and absolute convergence are, however, different concepts:

##### Example

**The alternating harmonic series**
\[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots\]
converges, but not absolutely.

(Idea) Draw a graph of the partial sums \((s_n)\) to get the idea that even and odd index partial sums \(s_{2n}\) and \(s_{2n+1}\) are monotone and converge to the same limit.

The sum of this series is \(\ln 2\), which can be derived by integrating the formula of a geometric series.

points are joined by line segments for visualization purposes

### Convergence tests

#### Comparison test

The preceeding results generalize to the following:

##### Theorem 4.

**(Majorant)** If \(|a_k|\le p_k\) for all \(k\) and
\(\sum_{k=1}^{\infty} p_k\) converges,
then also \(\sum_{k=1}^{\infty} a_k\) converges.

**(Minorant)** If \(0\le p_k \le a_k\) for all \(k\) and \(\sum p_k\) diverges,
then also \(\sum a_k\) diverges.

**Proof for Majorant.** Since \[a_k=|a_k|-(|a_k|-a_k)\] and
\[0\le |a_k|-a_k \le 2|a_k|,\]
then \(\sum a_k\) is convergent as a difference of two convergent positive series.
Here we use the elementary convergence property (Theorem 2.) for positive series;
this is not a circular reasoning!

**Proof for Minorant.** It follows from the assumptions that the partial sums of \(\sum a_k\)
tend to infinity, and the series is divergent.

\(\square\)

##### Example

Study the convergence of \[ \sum_{k=1}^{\infty} \frac{1}{1+k^3} \ \text{ ja }\ \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}. \]

**Solution.** Since
\[0<\frac{1}{1+k^3} < \frac{1}{k^3}\le \frac{1}{k^2}\]
for all \(k\in \mathbb{N}\), the first series is convergent by the majorant principle.

On the other hand, \[\displaystyle{\frac{1}{\sqrt{k}}\ge \frac{1}{k}}\] for all \(k\in\mathbb{N}\), so the second series has a divergent harmonic series as a minorant. The latter series is thus divergent.

#### Ratio test

In practice, one of the best ways to study convergence/divergence of a series is
the so-called **ratio test**, where the terms of the sequence are compared
to a suitable geometric series:

##### Theorem 5a.

Suppose that there is a constant \(0< Q < 1\) so that \[ \left| \frac{a_{k+1}}{a_k} \right| \le Q\] starting from some index \(k\ge k_0\).

Then the series \(\sum a_k\) converges (and the rate of convergence is comparable to the geometric series \(\sum Q^k\), or is even higher).

We may assume that the inequality is valid for all indices \(k\), because the initial part has no effect on the convergence (although it has an effect to the sum!).

This now implies that
\[|a_{k}|\le Q|a_{k-1}|\le Q^2|a_{k-2}|\le \dots\le Q^k|a_0|,\]
so the series has a convergent geometric majorant.

\(\square\)

#### Limit form of ratio test

##### Theorem 5b.

Suppose that the limit \[\lim_{k\to \infty} \left| \frac{a_{k+1}}{a_k} \right| = q\] exists. Then the series \(\sum a_k\) \[ \begin{cases}\text{converges} & \text{ if } 0\le q< 1,\\ \text{diverges} & \text{ if } q > 1,\\ \text{nay be convergent or divergent} & \text{ if } q=1. \end{cases} \]

**(Idea)** For a geometric series the ratio of two consecutive terms is
exactly \(q\). According to the ratio test, the convergence of some other
series can also be investigated in a similar way, when the exact ratio
\(q\) is replaced by the above limit.

In the formal definition of a limit \(\varepsilon =(1-q)/2>0\). Thus starting from some index \(k\ge k_{\varepsilon}\) we have \[ |a_{k+1}/a_k| < q + \varepsilon = (q+1)/2 = Q < 1, \] and the claim follows from Theorem 4.

In the case \(q>1\) the general term of the series does not go to zero, so the series diverges.

The last case \(q=1\) does not give any information.

This case occurs for the harmonic series (\(a_k=1/k\), divergent!) and superharmonic
(\(a_k=1/k^2\), convergent!) series. In these cases the convergence or divergence
must be settled in some other way, as we did before.

\(\square\)

##### Example

Is the series \[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}k}{2^k}= \frac{1}{2}-\frac{2}{4}+\frac{3}{8}-\dots\] convergent?

**Solution.** Here \(a_k=(-1)^{k+1}k/2^k\), so
\[
\left| \frac{a_{k+1}}{a_k}\right|
= \left| \frac{(-1)^{k+2}(k+1)/2^{k+1}}{(-1)^{k+1}k/2^k}\right|
=\frac{k+1}{2k} =\frac{1}{2}+\frac{1}{2k}\to \frac{1}{2} < 1,
\] when \(k\to\infty\). By the ratio test the series is convergent.