Convergence

So far we have derived power series formulae for standard functions (examples are shown above). We can also write down a function in the form of a series. Following Eq. (6.1), we have

which represents a function of $x$ with $c_{n}$ denoting an infinite list of coefficients. For instance, $c_{n}=1 / n$ ! leads to the exponential function. Coefficients of the form $c_{n}=(-1)^{m} 4^{-m} /(m !)^{2}$ lead to Bessel functions; these functions are of the nature of decaying sinusoidal waves which have practical applications in many fields. The function defined by (6.31) is understood as the limit,

Note that (6.32) represents the limit of the partial sums of the series. If the limit exists and it is finite, the series is called convergent. If the limit does not exist or if the sequence alternates between $\infty$ and $-\infty$, we refer to the series as divergent.

Examples of convergent and divergent series

This chapter is mostly concerned about power series which take the form of (6.1). In what follows, for simplicity, we only use numbers in the series. In particular, we use the series $\sum_{n} a_{n}$ and $a_{n}$ is referred to as the series term. This is also the case in Subsec. 6.2.2 where we discuss tests for convergence. Note that the fact that we use numbers and not variables in our discussion here does not change the results for power series; everything we learn here applies to power series as well. We return to power series in Subsec. 6.2.3.

# 1: Consider the series,

the general formula ${ }^{10}$ for the partial sums is given by

Next, we compute the limit as in (6.32),

# 2: The following series is known as the harmonic series,

The series (6.34) diverges - see section on integral test for proof.

# 3: As a final example consider,

The sequence of partial sums is

thus oscillating from 0 to 1 . The sequence, $S_{n}$ diverges since $\lim _{n \rightarrow \infty} S_{n}$ does not exist; therefore the series given by (6.35) also diverges.

Absolute convergence

Absolute convergence is a stronger type of convergence.

Theorem (Absolute convergence)

If $\sum_{n}\left|a_{n}\right|$ converges, then so does $\sum_{n} a_{n}$ and we say that $\sum_{n} a_{n}$ converges absolutely ${ }^{11}$.

${ }^{10}$ This is an example of a telescoping series; its partial sums yield only a fixed number of terms after cancellation. By expressing the series as partial fractions and expanding out a few terms of the series allows us to see this result easily.

${ }^{11}$ If $\sum_{n} a_{n}$ is convergent and $\sum_{n}\left|a_{n}\right|$ is divergent, we refer to the series as conditionally convergent. To see this, we write

since $\left|a_{n}\right|$ is either $a_{n}$ or $-a_{n}$. We assume that $\sum_{n}\left|a_{n}\right|$ is convergent and so $\sum_{n} 2\left|a_{n}\right|$ is also convergent since we can simply factor the 2 out of the series. Using the comparison test (details in Subsec. 6.2.2) and Eq. (6.36), we know that if the larger series, i.e. $\sum_{n} 2\left|a_{n}\right|$, converges then so does the smaller one, i.e. $\sum_{n} a_{n}+\left|a_{n}\right|$. This allows us to write,

thus, $\sum_{n} a_{n}$ is the difference of two convergent series and so it is also convergent.

Standard tests

We have now seen examples of both convergent and divergent series by essentially evaluating the series. This will generally not be an easy task so usually we resort to standard tests to help us identify whether a series is convergent or divergent.

Divergence test

This is also referred to as the preliminary test and it tells us whether a series diverges. Before we get into the details of the test, we note a very important limitation: the divergence test determines whether a series diverges and if it does diverge then it cannot converge but the divergence test is not a test for convergence. More particularly, it is possible for a series to pass the divergence test such that the series appears to converge but still diverge when another test is applied. An example is

The divergence test is stated as follows. If $a_{n} \nrightarrow 0$ as $n \rightarrow \infty$, then $\sum_{n} a_{n}$ diverges. ${ }^{12}$ Note that this statement is equivalent to the contrapositive: if $\sum_{n} a_{n}$ converges, $a_{n} \rightarrow 0$ as $n \rightarrow \infty$.

To show this, consider the partial sums of the series,

where the terms in the brackets make up the quantity $S_{n-1}$ such that, when rearranged, we have

We now use the contrapositive statement above: if $\sum_{n=1}^{\infty} a_{n}=L$ (i.e. converges) then

${ }^{12}$ The notation ' $a_{n} \nrightarrow 0$ as $n \rightarrow \infty$ ' is valid when the limit $\lim _{n \rightarrow \infty} \neq 0$ or when the limit does not exist. Here, we note that $n-1 \rightarrow \infty$ as $n \rightarrow \infty$ so the limits are the same.

Integral test

We use the example of the harmonic series [see Eq. (6.34)] to show that it diverges using the integral test. Suppose $f$ is a continuous, positive, decreasing function defined on $[1, \infty)$ and let $a_{n}=f(n)$. The series $\sum_{n=1}^{\infty} a_{n}$ is convergent if and only if the improper integral

is convergent. By the integral test, we have the following result:

1. if $\int_{1}^{\infty} f(x) d x$ is convergent, then $\sum_{n=1}^{\infty} a_{n}$ is convergent;
2. if $\int_{1}^{\infty} f(x) d x$ is divergent, then $\sum_{n=1}^{\infty} a_{n}$ is divergent.

Note on improper integrals

Consider the following integral

where the function $1 / x$ is continuous, positive, and decreasing. The integral is defined over an infinite interval; we proceed by converting the integral to a limit by replacing the upper integral limit of infinity by a variable $t$ and take the limit as $t$ tends to infinity,

Carrying out the integral and the limit gives

Using the integral test therefore with $f=1 / x$, we conclude that since the improper integral above [Eq. (6.37)] is divergent, so is the series given by Eq. (6.34).

General remarks

• Do not use the integral test to evaluate the series since, in general,

The test only determines the convergence of a series, not its value.

• The series does not need to start at $n=1$ for the integral test to be valid.
• With regard to the function $f$ be decreasing: $f(x)$ does not need to be always decreasing. It should be ultimately decreasing, that is to say that it should be decreasing for $x$ larger than some number $N$.

Comparison test

The integral test requires us to carry out the improper integral which is not always an easy task. The comparison test works by comparing the series we are interested in to one whose convergence properties are known. The test is stated as follows. Suppose that $\sum a_{n}$ and $\sum b_{n}$ are series with positive terms. If:

1. $\sum_{n} b_{n}$ is convergent and $a_{n} \leq b_{n}$ for all $n$, then $\sum_{n} a_{n}$ is also convergent.
2. $\sum_{n} b_{n}$ is divergent and $a_{n} \geq b_{n}$ for all $n$, then $\sum_{n} a_{n}$ is also divergent.

The test states that if we have two series of positive terms and the terms of one are always larger than the terms of the other then, if the larger series is convergent so is the smaller one. Similarly, if the smaller series diverges, so does the larger one.

For instance, consider the series

The idea of the comparison test is to first guess whether the series converges (diverges) and then find a sequence that is larger (smaller) than the one we have which we know to converge (diverge). Note that, in the case of (6.38), for large $n$,

and we know $\sum_{n=1}^{\infty} \frac{1}{n}$ to diverge (from the integral test) so we guess that the series diverges. To find a smaller sequence, we start off by making the denominator of the term in series (6.38) bigger. Since $n \geq \sqrt{n}$ for $n \geq 1$, then $2 n \geq n+\sqrt{n}$ from which it follows that

Finally, since $\sum_{n=1}^{\infty} \frac{1}{2 n}$ diverges, $\sum_{n=1}^{\infty} \frac{1}{n+\sqrt{n}}$ also diverges.

Ratio test

We can use the ratio test to tell us whether a series is absolutely convergent or divergent. Since absolute convergence implies convergence as well (see Subsec. 6.2.1), the ratio test is often used to determine the convergence of a series. Suppose we have a series $\sum_{n} a_{n}$. Let

then, we have:

1. if $L<1$, the series is absolutely convergent;
2. if $L>1$, the series is divergent;
3. if $L=1$, the test is inconclusive. The extension to the inconclusive case is given by Raabe's test (see also Example 6.4). That is, if for the case $L=1$, we can further show that

then we can say that the series converges absolutely if $L^{\prime}>1$ and diverges if $L^{\prime}<1$. If $L^{\prime}=1$, we again obtain no information.

Note that in the case where we obtain no information, the series may be absolutely convergent, conditionally convergent, or divergent. We consider some examples next.

Example 6.3 Consider the series

use the ratio test to show whether the series converges or diverges.

Solution To use the ratio test, we first need to compute $L$ :

Since $L=0$, the series converges by the ratio test.

The ratio test is very powerful with respect to power series. Note however that the test is always inconclusive for series of the form $\sum \frac{\text { polynomial }}{\text { polynomial }}$. Example 6.4 Consider the $p$-series given by,

whose sum (as a function of $p$ ) is known as the Riemann zeta function. Apply the ratio test to determine its convergence.

Solution Applying the ratio test

which yields no information. Note that we can write

as

and so, expanding beyond the leading term,

where we have used the binomial theorem. Applying Raabe's test [Eq. (6.40)], we observe that (6.43) converges for $p>1$ and diverges for $p<1$. For $p=1$, we obtain no information but in this case, (6.43) is the harmonic series [see Eq. (6.34)] which we know to diverge.

Theorem (Linearity)

provided that both series on the RHS converge.

Consider the following series,

It is tempting to split (6.46) into $\sum 1 / n-\sum 1 /(n+10)$ and claim that it diverges because the two series do. But the theorem [Eq. (6.45)] endorses this procedure only if both series on the right-hand-side of Eq. (6.45) converge, and here they do not. Instead note that

and since $\sum b_{n}$ is a convergent series (refer to Example 6.4 with $p=2$ ), it follows that $\sum a_{n}$ converges also.

Alternating series test

An alternating series is any series $\sum_{n} a_{n}$ which can be written in the form

where $b_{n} \geq 0$. If the sequence $\left\{b_{n}\right\}$ satisfies the following conditions:

1. $b_{n} \geq 0$ for sufficiently large $n$,
2. $b_{n+1} \leq b_{n}$ for sufficiently large $n$ (i.e. $\left\{b_{n}\right\}$ is a decreasing sequence),
3. $b_{n} \rightarrow 0$ as $n \rightarrow \infty$,

then $\sum_{n} a_{n}$ converges.

General remarks

• The test applies when $a_{n}$ takes the form $a_{n}=(-1)^{n} b_{n}$;
• The test can only be used to determine if a series is convergent and not if it diverges;
• In the second condition above, we require that $b_{n}$ eventually decreases; that is to say that while the first few terms may be increasing, we need $b_{n+1} \leq b_{n}$ for all $n$ after some number $N$. This is easy to see: consider $\sum_{n=1}^{\infty}(-1)^{n} b_{n}$; suppose for $1 \leq n \leq N$ the sequence $b_{n}$ is not decreasing and for $n \geq N+1$ it is decreasing. We can then write

The first series on the RHS has a finite value which we can compute. The second series on the RHS, i.e. the infinite series, determines the convergence of the original infinite series. If the second series has a finite value, then the sum of two finite series is also finite and so the original series converges. So the sequence $b_{n}$ does not need to be decreasing for all $n$, it is only required that the sequence is eventually decreasing.

As an example, consider the series,

The series is referred to as the alternating harmonic series. We identify $b_{n}$ in (6.47) as $b_{n}=1 / n$. To determine if the series is convergent we test whether the conditions outlined above are met:

since

and $\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} 1 / n=0$. Hence, all conditions are met and by the alternating series test, the series converges.

Theorem (Dirichlet test)

The alternating series test is a particular case of the Dirichlet test which gives a method to test the convergence of a series. If $b_{n} \downarrow 0^{13}$ as $n \rightarrow \infty$ and the sequence $a_{n}$ has bounded partial sums, then the series $\sum a_{n} b_{n}$ converges.

Let us take the alternating series (6.48) with $1 / n$ to be $b_{n}$ and $(-1)^{n+1}$ to be $a_{n}$. Then $1 / n \downarrow 0$ as required and the partial sums

are surely bounded. For instance

and the series is convergent. The Dirichlet test implies that any alternating series $\sum(-1)^{n+1} b_{n}$ converges if $b_{n} \downarrow 0$. Of course, not all series with terms of mixed sign are alternating series.

Radius of convergence

We now return to power series [refer to Eq. (6.1)] which, as seen in Section 6.1, involve a variable, $x$. A power series may converge for some values of $x$ and not others. We show that there is a nonnegative number $R$ so that the power series converges for

and diverges for

for $\left|x-x_{0}\right|=R$, the series may or may not converge. This number is called the radius of convergence for the series while the interval of all $x$ values for which the series is convergent is called the interval of convergence of the series. The name 'radius' originates from complex power series (covered in Chapter 7) where the condition $|z|<R$ with $z$ the complex number $\mathrm{z}=\mathrm{x}+\mathrm{i} \mathrm{y}$ is a circle of radius $R$ centred at the origin. In general, if a power series is expanded around a point $a$ and the radius of convergence is $R$, then the set of all points $z$ such that $|z-a|=R$ is a circle called the boundary of the disk of convergence. A power series may diverge at every point on the boundary, or diverge on some points and converge at other points, or converge at all the points on the boundary. After we establish the radius of convergence of a series we must check what happens at the boundary, or endpoints if we are on the real axis.

We note that power series always converge for at least one value, that is $x=x_{0}$. Taking $x=x_{0}$ in Eq. (6.1)

since all terms beyond $a_{0}$ are multiplied by $\left(x_{0}-x_{0}\right)$. If the series converges only at $x=x_{0}$ then $R=0$ and if the series converges for all values of $x, R$ is said to be infinite. To determine the radius of convergence for a series, we need to use one of the tests discussed

${ }^{13}$ By $b_{n} \downarrow 0$ as $n \rightarrow \infty$, we mean that $b_{n}$ is monotone decreasing to zero; that is, $m>n$ implies that $b_{m} \leq b_{n}$ and $b_{n} \rightarrow 0$ as $n \rightarrow \infty$. above (usually, the ratio test is used). Let us look at an example. The power series for the function $f(x)=1 /(1-x)$, expanded around $x=0$ is simply

We know that the series converges at $x=0$ (the series is about the point $x_{0}=0$ ); we apply the ratio test [Eq. (6.39)] to determine for what $x$ values the series converges/diverges. Upon application of the test, we determine the radius of convergence as well as the interval of convergence. As mentioned above, the endpoints may need to be separately tested to observe whether the interval is open (endpoints not included) or closed (endpoints included).

Using the ratio test,

According to the ratio test if $R<1$, the series is absolutely convergent, if $R>1$, it diverges; it follows that for $|x|<1$, the series is absolutely convergent and for $|x|>1$, the series diverges. The radius of convergence is $R=1$.

Finally, we ask whether the power series converges or not at the endpoints, i.e. at $x= \pm 1$. The cases $x=+1$ and $x=-1$ must be examined separately, and it is easily seen that in both cases the series diverges. We conclude that the series converges absolutely for $|x|<1$ and diverges for $|x| \geq 1$. The interval of convergence is therefore $(-1,1)^{14}$.

Alternating series revisited

The power series for $f(x)=-\ln (1-x)$, expanded around $x=0$, given by

has radius of convergence 1 (using the ratio test) and diverges for $x=1$ as it yields the series

which is the harmonic series in (6.34) which we know to be divergent. For $x=-1$ the series becomes the alternating series

which is the negative of the alternating harmonic series, Eq. (6.48)

If $\sum_{n}\left|a_{n}\right|$ converges, then we know from Subsec. 6.2.1 that $\sum_{n} a_{n}$ also converges (absolute convergence). But

${ }^{14}$ Note that the notation $(-1,1)$ implies that neither endpoint is included in the interval (denoted by round brackets). Contrast this with $[-1,1)$ which implies that the left point i.e. -1 is included (denoted by a square bracket) while the right endpoint, 1 , is not included (denoted by a round bracket) is the harmonic series which is divergent; so absolute convergence does not apply. Using the alternating series test, we showed that the series (6.48) converges. The same result was also concluded with the Dirichlet test. The series converges conditionally since it is convergent but not absolutely convergent (from footnote 11).

Let us now prove that the alternating series (6.52) converges to $\ln 2$ (note that we started with the function $f(x)=-\ln (1-x)$ but we took the negative of the resulting series for $x=-1$ when moving from Eq. (6.51) to (6.52) hence we proceed to prove that series (6.52) converges to $\ln 2)$. Consider the Taylor expansion of $\ln (1+x)$ about $x=0$,

Use the ratio test to determine the radius of convergence,

and the series converges absolutely for $|x|<1$ and $R=1$. Next, we check the end points: for $x=-1$ the series becomes $-1-\frac{1}{2}-\frac{1}{3}-\cdots$ which is divergent. For $x=+1$ the series becomes $1-\frac{1}{2}+\frac{1}{3}-\cdots$, which is the alternating power series in (6.52) which is convergent. Hence, the series converges in $(-1,1]$ or for $|x|<1$ and $x=1$. By comparison now with (6.52),

As a final example on absolute convergence, consider the power series

which has radius of convergence 1 (ratio test) and converges absolutely for $|x|<1$ but also on the boundary $|x|=1$. Hence the series converges absolutely for $|x| \leq 1$.

Types of series

We briefly outline two basic types of series which are useful to know when finding whether a series is convergent or divergent.

$p$-Series

The $p$-series (also seen in Example 6.4) takes the form

and it is convergent if $p>1$, divergent if $p \leq 1$.

Geometric Series

The geometric series,

is convergent if $|r|<1$ and divergent if $|r| \geq 1$.

Section 6.2 exercises

1. Determine whether the following series converges or diverges. If the series converges, determine its sum.

2. Using the integral test, determine whether the following series is convergent or divergent,

3. Determine if the following series converges or diverges using the comparison test,

4. Compute $L=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|$ and then use the ratio test to determine if the series converges or diverges.

5. Find the radius and interval of convergence of the series,

6. Investigate the convergence of the following series,