Series (mathematics)

In mathematics, a series is the cumulative sum of a given sequence of terms. Typically, these terms are real or complex numbers, but much more generality is possible.

For example, given the sequence of the natural numbers 1, 2, 3, ..., the series is
1,
1 + 2,
1 + 2 + 3, ...
The above writing stresses the 'cumulative' nature of the series and is justified by the mathematical definition we introduce below, but more direct notation is typically used: 1 + 2 + 3 + ... .

Depending on the number of terms, the series may be finite or infinite. The former is relatively easy to deal with. The finite series is identified with the sum of all terms and — apart from the elementary algebra — there is no particular theory that applies. It turns out, however, that much care is required when manipulating infinite series. For example, some simple operations borrowed from the elementary algebra — such as a change of order of the terms — often lead to unexpected results. So it is sometimes tacitly understood, especially in analysis, that the term "series" refers to the infinite series. In what follows we adopt this convention and concentrate on the theory of the infinite case.

Motivations and examples

Fig. 1. Graphical representation of a geometric series.

Given a series, an obvious question arises. Does it make sense to talk about the sum of all terms? That is not always the case. Consider a simple example when the general term is constant and equal to 1, say. That is, the series reads as 1+1+1+1+... (without end). Clearly, the sum of all terms can not be finite. In mathematical language, the series diverges to infinity. There is not much to say about such an object. If we want to build an interesting theory, that is to have some examples, operations and theorems, we need to deal with convergent series, that is series for which the sum of all terms is well-defined.

Actually, are there any? Maybe any series would diverge like this? Consider the following series of decreasing terms, where every term is half the previous term:

${\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+{\frac {1}{32}}+\ldots$

This is a special example of what is called geometric series. The sum is finite! Instead of a rigorous proof, we present a picture (see fig.1) which gives a geometric interpretation. Each term is represented by a rectangle of the corresponding area. At any moment (given any number of rectangular "chips on the table"), the next rectangle covers exactly one half of the remaining space. Thus, the chips will never cover more than the unit square. In other words, the sum increases when more terms are added, but it does not go to the infinity, it never exceeds 1. The series converges.

One may tend to think that any decreasing sequence of terms would eventually lead to a convergent series. This, however, is not the case. Consider for example the series

$1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+\ldots$

To see that it diverges to infinity, let us make the following groups of terms (we may and do forget the one at the beginning):

1/2,

1/3+1/4,

1/5+1/6+1/7+1/8,

1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/16,

etc. We will show that sum of each line is at least 1/2. Notice that each group ends by a term of the form $1/2^{n},$ the smallest one in the group. Note also that group that ends by $1/4$ has two members, the one that ends by $1/8$ has four members and so on. Generally, a group that ends by $1/2^{n}$ has $2^{n-1}$ terms, each of which is greater than the smallest one at the end. So pick a group and replace each term by the smallest one. Then the new sum is easy to calculate:

$2^{n-1}$ members $\times {\frac {1}{2^{n}}}$ (this is the smallest term) $=1/2$

Obviously, the original sum is at least as big as this and our claim follows: the sum of each group is at least one half. And since there are infinitely many of such groups, the sum of all terms is not finite.

These examples show that we need criteria to decide whether a given series is convergent. Before such tools can be developed we need some mathematical notation.

Formal definition

Given a sequence $a_{1},a_{2},\dots$ of elements that can be added, let

$S_{n}=a_{1}+a_{2}+\cdots +a_{n},\qquad n\in \mathbb {N} .$

Then, the series is defined as the sequence $\{S_{n}\}_{n=1}^{\infty }$ and denoted by $\sum _{n=1}^{\infty }a_{n}.$ For a single n, the sum $S_{n}$ is called the partial sum of the series.

If the sequence $(S_{n})$ has a finite limit, the series is said to be convergent. In this case we define the sum of the series as

$\sum _{n=1}^{\infty }a_{n}=\lim _{n\to \infty }S_{n}.$

Note that the sum (i.e. the numeric value of the above limit) and the series (i.e. the sequence $S_{n}$) are usually denoted by the same symbol. If the above limit does not exist - or is infinite - the series is said to be divergent.

Series with positive terms

The simplest to investigate are the series with positive terms $(a_{n}\geq 0).$ In this case, the cumulative sum can only increase. The only question is whether the growth has a limit. This simplifies the analysis and results in a number of basic criteria. Notice that it is not positivity that really matters. If a given series has only negative terms, a "symmetric" argument can always be applied. So the only thing we really need is the constant sign. However, for the sake of clarity, in this section we assume that the terms are simply positive.

Comparisons

There is a "family" of relative tests. The nature of a given series is inferred from what we know about another series, possibly simpler one.

If for two series $\sum a_{n}$ and $\sum b_{n}$ we have $a_{n}\leq b_{n}$ and if the series $\sum b_{n}$ is known to be convergent, then the other one, $\sum a_{n},$ is convergent as well. In other words a series "smaller" than a convergent one is always convergent. This is not difficult to justify. When we consider series of positive terms, the cumulative sum may only increase. For convergent series the growth is not unlimited -- there is an upper bound. Accordingly, the growth of the "smaller" series is limited by the same bound.

The above argument works the other way round too. If for the two series we have $a_{n}\geq b_{n}$ and the series $\sum b_{n}$ is known to be divergent, then the other one, $\sum a_{n},$ diverges as well. In other words a series "bigger" than a divergent one is divergent too.

A very useful version of comparison test may be expressed as follows. If for two series $\sum a_{n}$ and $\sum b_{n}$ it can be established that

$\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c$

with a finite strictly positive constant $c\in (0,\infty )$, then both series are of the same nature. In other words, if one series is known to converge, the other converges as well; if one diverges, so does the other.

Comparison with integral. This can be done if the sequence of general terms is decreasing, that is $a_{n}>a_{n+1}.$ More precisely, suppose that we have $f(n)=a_{n},\,n=1,2,...,$ for a certain decreasing function f defined on $(1,\infty )$. If the integral

$\int _{1}^{\infty }f(x)\,dx$

converges, so does the series $\sum a_{n}.$

This equivalence is quite important since it allows to establish convergence of a given series using basic methods of calculus. Prominent examples include the Riemann series, that is the series with general term $a_{n}=1/n^{p}$ with a constant $p\in (0,\infty ).$ If we take the function $f(x)=1/x^{p}$ then we get

$\int _{1}^{\infty }f(x)\,dx=[\ln x]_{1}^{\infty }$ if $p=1$ and

$\int _{1}^{\infty }f(x)\,dx=\left[{\frac {x^{-p+1}}{-p+1}}\right]_{1}^{\infty }$ otherwise.

The limit at infinity exists if and only if $p>1$. It follows that the series

$\sum {\frac {1}{n^{p}}}$

converges if and only if $p\in (1,\infty ).$

Absolute tests

There are also tests that allow to determine the nature of a given single series. Two most popular are listed below.

• D'Alembert ratio test

In its simplest form it involves computation of the following limit

$L=\lim _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}.$

If L is strictly greater than 1, then the series diverges. If L < 1 then the series is convergent. The case L = 1 gives no answer. Indeed, take for example $a_{n}=1/n$. A short computation gives L = 1. Notice that we showed that this series is divergent. On the other hand, if $a_{n}=1/n^{2}$, the series was shown to converge. Still, if we compute L from the above formula, we get L = 1.

• Cauchy root test

In its popular form it is based on the computation of the following number

$L=\limsup _{n\to \infty }{\sqrt[{n}]{a_{n}}}$

($\limsup$ here refers to the upper limit of the sequence). Similarly as before, L > 1 implies divergence, L < 1 means that the series is convergent. The case L= 1 does not allow to decide.

Which one of the two above tests is more appropriate depends on the form of the general term. However, not only computational convenience is in question. In some cases only the root test can be applied. Consider for example the series 1 + 1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/16 + 1/16 +.... Formally, the limit of the ratio $a_{n}/s_{n+1}$ does not exist (the upper limit is equal to 1, the lower limit is 1/2). Nonetheless the root test gives $L=1/{\sqrt {2}}$ and this determines that the series converges.

In general, the root test is more universal in the following sense: if the limit L of the ratios $a_{n+1}/a_{n}$ exists, then the limit of roots ${\sqrt[{n}]{a_{n}}}$ exists as well and both are equal. On the other hand, if the ratio test fails to give an answer due to L = 1, then there is no hope that the root test allows to decide.

Series with arbitrary terms

Fig. 2 Alternating series converges if the "steps" are shortening.

In general, a given series may contain terms of different sign. This may lead to very delicate situations. Consider a simple example, called the alternating harmonic series

$1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\ldots .$

Denote the general term by $a_{n}=(-1)^{n}/n.$ Imagine that each term represents a small step along a line. Considering the sign, we swing back and forth. And it is easy to see that due to shortening the step we will approach a certain point, just like a damped pendulum -- see Fig. 2. This informally shows that $\sum a_{n}$ converges. Furthermore, we know the actual sum of this series, it equals to $\ln 2$ (this needs some actual computations, though). Now we reorganise the terms and consider

$1-{\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{3}}-{\frac {1}{6}}-{\frac {1}{8}}+\ldots ..$

The general term reads

$b_{n}={\frac {1}{2n+1}}-{\frac {1}{2(2n+1)}}-{\frac {1}{2(2n+2)}},\quad n=0,1,2,\ldots$

Clearly, this series contains the same terms as $\sum a_{n}$ just in different order. Furthermore, simplifying the expression for $b_{n}$ gives

$b_{n}={\frac {1}{2(2n+1)}}-{\frac {1}{2(2n+2)}}={\frac {1}{2}}\left({\frac {1}{2n+1}}-{\frac {1}{2n+2}}\right)$

so that

$b_{0}+b_{1}+b_{2}+\ldots ={\frac {1}{2}}\left[(1-{\frac {1}{2}})+({\frac {1}{3}}-{\frac {1}{4}})+\ldots \right].$

It follows that the total sum is $1/2\cdot \ln 2$. And we changed only the order...

How is it possible? It turns out that the reason is that the series with the modulus $\sum |a_{n}|$ does not converge at all (see the second example in the "motivations" section above). Even more strange general result can be shown: for any given number $M\in \mathbb {R}$ we can find a rearrangement of terms in our series $\sum a_{n}$ to get the total sum of M.