Thus,
We see that 
is a number with the required
property. Real Numbers and the Concept of Limit
In Volume I, 1.1, it was taken for granted that the real numbers form an aggregate within which the ordinary operations of arithmetic may be performed as with the rational numbers. We shall now examine this assumption more closely. We take the arithmetical operations on the rational numbers as given. Our objective is then to make an abstract analytical extension of the class of rational numbers which shall yield the wider class of real numbers, and to do this without relying on intuition in our proofs. We must frame our definitions in such a way that, as a logical consequence of them, the ordinary rules of arithmetic apply to all real numbers just as they do to rational numbers.
The introduction of irrational numbers will be undertaken in close conjunction with a thorough consideration of the concept of limit, in which we shall repeat in a revised form the discussion of Volume I, Appendix 1.
The only difference in the point of view will be that here we shall start with the logical, abstract concept of real numbers, while formerly the properties of real numbers were taken for granted.
1. Definition of the Real Numbers by means of Nests of Intervals:.The irrational numbers and, in general, the real numbers were defined in Volume I, 1.2, by means of decimals, the rational numbers being represented by terminating or recurring decimals. By such a decimal, say a = 0.a1a2a3 ··· ,we mean that the number represented, called a, lies between the rational number an = 0.a1···an and the rational number an + 10-n. The number a is thus determined by means of a sequence or nest of progressively smaller and smaller intervals, each inside the previous one, the n-th interval being of length 10-n.
For our present purpose, it would be inconvenient to restrict ourselves to special nests of intervals, where the length of the n-th interval is 10-n. We begin with the following general definition.
We mean by a rational interval (a|b) the aggregate of all the rational numbers x which satisfy the inequalities a £ x £ b, where a < b and a and b are rational numbers. The number (b - a) is called the length of the interval. We say that the interval (c|d) is contained in the interval (a|b) if a £ c <d £ b. An infinite sequence of rational intervals (a1|b1), (a2|b2) - - . is called a nest of intervals if every interval (an|bn) contains the next - (an+1|bn+1), and the lengths bn - an tend to zero, i.e., given any positive number e, however small (the number must, of course, be rational, since no other numbers have as yet been introduced), there is a number N(e) such that the lengths bn - an are less than e for all subscripts n which exceed N.
We arrive from the intuitive meaning of a nest of intervals and remembering,. in particular, how we may pick out any point on the number axis by means of a nest of intervals, as in Volume I, 1.2, at the idea that we may define an arbitrary real number by a nest of intervals. This is to be taken as meaning the following: The real number is given by an unending process of approximation which is determined by the nest of intervals. The nest with the general member (an|bn) gives us, with regard to the number a to be defined, the fact that this real number lies between a1 and b1; again, it lies between a2 and b2 , etc. The nest of intervals will thus yield two rational numbers, as close to each other as we please, between which lies the real number
The essential step is now that we abandon the notion of obtaining an objective definition of the irrational numbers. We give up the attempt to characterize the irrational numbers as given mathematical entities with specific properties. We do not say that an irrational number is such and such a mathematical object; instead, we are content with the process of approximation which gives the nest of intervals and regard each such process as the definition of a real number. If there is a rational number a contained in all the intervals (an|bn ), the real number defined by the nest of intervals (an|bn ) is said to be identical with a. By this assumption, the rational numbers become also real numbers.
The words irrational number or, more generally, real number may thus be regarded merely as an abbreviated reference to a nest of intervals.
Some process of this kind is often essential in giving a precise formulation to mathematical concepts. For instance, in projective geometry, when points at infinity are introduced, these points are not treated as definite mathematical entities in themselves; we merely say that a point at infinity is given by a pencil of parallel lines.
This is what is meant by the statement that an irrational number is given or defined by a nest of intervals. In practice, it comes to this, that every operation with real numbers is an operation with nests of intervals. This offers the possibility of making calculations with real numbers depend logically on operations with rational numbers.
It is necessary to lay down a procedure for defining addition, multiplication, etc. of real numbers by nests of intervals. Here the rules must be framed in such a way that the ordinary laws of calculation still apply. Moreover, we must ensure that the rules of calculation with rational numbers are not contradicted.
We shall begin by showing that our definition implies an ordering of the real numbers by magnitude. This in itself provides sufficient groundwork for the axiomatic foundation of the concept of limit and a more thorough understanding of it. When this has been achieved, we shall return to the question of the rules of calculation with real numbers.
2. The Real Numbers in Order of Magnitude: Let two numbers a and g be given by nests of intervals (an|bn) = in and (cn|dn) = jn. The following three cases may arise:
(1) From a certain stage n = n0 onward every interval jn lies to the right of the interval in, i.e., for n = n0, and, of course, for every n > n0, we have bn < cn. We then say that g is larger than a or g > a.
(3) Neither of the above situations arises. We then say that the two nests of intervals in and jn define the same number a = g: Thus, two nests of intervals define the same number if, and only if, the intervals in and jn always overlap, i.e., if both an £ dn and bn £ cn; or, if the two intervals in and jn have rational points in common for every n. A special consequence of this definition is that, if of two nests of intervals the one is obtained from the other by the omission of a finite or infinite number of constituent intervals, the two nests define the same real number.
All these rules, giving the magnitude relations between two real numbers, can be understood immediately from the point of . view of the intuitive meaning of nests of intervals.
A few simple facts about inequalities between real numbers will now be noted. They will be of use in what follows.
We first make the following observation. The relation a £ g can be inferred from the two defining nests of intervals (an|bn) for a and (cn|dn) for g if we note that from a certain n = n0 onwards there holds the inequality a n £ g.
In fact, if a = g, this inequality is satisfied, as can be seen from the definition of equality, and if a < g then from some number onwards we have b n < cn so that a fortiori a n < dn. Conversely, if from some number onwards a n £ dn, then either bn³cn for all such values of n, and the a = g, by definition, or else, for some value of an, bn< cn which yields a < g.
In just the same way, we see that the condition c n £ bn for all large values of n is equivalent to a ³ g.
We see at once from the above that if a is a real number determined by the nest of intervals (an|bn) , then an £ a £ bn. This fact justifies our rule, because it shows that any real number is actually contained in every interval of the nest which defines it.
If a and b are two real numbers and a < b, then by the interval (a |b) is meant the aggregate of all real numbers x such that a £ x £ b. We call an interval a rational interval if its end-points a. and b. are rational numbers. We say that the real number x lies in the interior of the interval, if the signs of equality are absent, so that a < x < b. We describe (a |b) as a neighbourhood of the real number g if g lies in the interior of (a |b).
Every interval has rational numbers r in its interior.
In fact, let (an|bn)
and (cn|dn)
be nests of intervals defining the numbers a
and b . Since a <
b , there is a number n0 such that Thus,
We see that is a number with the required
property.
We obtain form this the following statement: If (a |b) is a neighbourhood of g, then (a |b) contains a rational neighbourhood (a|b) of g. It is only necessary to choose two rational numbers a and b such that a < a <g < b< b. It is also easy to see that if a < b, then rational neighbourhoods (a|b) of a and (c|d) of b can be found such that b < c; in other words, the two neighbonrhoods do not have points in common.
We shall not deal with the fundamental rules of calculation until we come to section 8, p. 680. Our next step is to resume the analysis of the concept of limit with the help of the ideas just explained.
3. The Principle of the Point of Accumulation: The determination of real numbers by nests of intervals forms the essential basis of the proof of the principle of the point of accumulation, which is due to Weierstrass. At first, a few remarks on the concept of the point of accumulation will first be made.
This discussion is essentially a repetition of earlier work, and so are the next 3 sections
Let M be an infinite set of real numbers in which it is permissible for the same number to occur more than once, and indeed an infinite number of times. (For example, 1, 1, 1, ··· is such a set.) If x is a number such that every neighbourhood of x contains an infinity of numbers belonging to M, then x is called a point of accumulation of the set M. Naturally, the name recalls the geometrical connection between numbers and points. Since every neighbourhood of x contains a rational neighbourhood, it is sufficient to formulate the above requirement in terms of rational neighbourhoods only.
An infinite set of numbers need not necessarily have a point of accumulation. The set of integers provides an example. A point of accumulation of a set need not itself be a member of the set. For example, the set 1, 1/2, 1/3, ··· , l/n, ··· has 0 as point of accumulation, but the definition of the set shows that 0 is not one of its members. A set which contains all its points of accumulation is said to be closed. The set of all numbers x such that 0 < x < 25 is not closed, since the points of accumulation 0 and 25 do not belong to it. On the other hand, 0 £ x £ b defines a closed set. A set a £ x £ b is called a closed interval.
A set may have an infinity of points of accumulation. For example, every real number is a point of accumulation of the set of rational numbers. In fact, if a is any real number, which may be thought of as given by a nest of intervals (a|b), then every neighbourhood of a contains an infinity of intervals (an|bn) and hence of rational numbers (an ,bn).
The principle of the point of accumulation, which will now be proved runs as follows: Every bounded infinite set of real numbers, i.e., every infinite set of real numbers lying in a definite interval, possesses at least one point of accumulation.
In order to prove this statement, we have to construct a nest of intervals defining a real number which has the property of a point of accumulation of the set. To start with, we observe that it is legitimate to assume that the given set is contained in a rational interval; in fact, if this were not the case, we could replace the given interval by a larger interval with rational end-points. We now subdivide this rational interval into two equal sub-intervals. At least one of these contains an infinite number of points of the set. In fact, if this is not the case, the original interval contains only a finite number of points of the set, and the hypothesis is contradicted. We take the sub-interval containing an infinite number of points of the set, or, if such occur in both, we take one or the other of the sub-intervals and subdivide it into two equal intervals. Just as before, at least one of these sub-intervals contains an infinity of points of the set. Either this one, or one of the two containing an infinity of points of the set, is now re-sub-divided, etc. In this way, a nest of intervals (an ,bn) is constructed, since each interval taken is contained in the previous one and the length of the n-th interval is one 2-n-th part of the length of the original interval. This nest of intervals defines a real number x. Next, we will show that x is a point of accumulation of the set.
Consider any rational
neighbourhood (r| s) of x , so that r < x < s. Then,
from a certain number n1 onward, we must have
and from another (possibly different) number n2
onward 
In any case, if n > n1,
and also n > n2, then (an,bn)
is contained in (r|s). The construction of our nest (an,bn)
shows that each interval of the nest contains an infinity of
points of the set, whence the arbitrary rational neighbourhood (r|s)
of x also contains an
infinity of points of the set. But this asserts precisely the
fact that x is a point of
accumulation of the set.
4. Upper and Lower Points of Accumulation. Upper and Lower Limits: In the construction, which has just led us to a point of accumulation of a bounded infinite set, we might have introduced the restriction that the second interval (that with the larger numbers as its end-points) should always be chosen whenever it contained an infinity of points of the set. If this were done, the nest of intervals obtained would define a perfectly definite point of accumulation b of the set. This number b is the largest of the numbers corresponding to points of accumulation of the set.
This follows at once from the remark that there can only be a finite number of points of the set in any interval to the right of each interval (an,bn) of the nest described above.
If g is an arbitrary number
larger than b and if it is
sufficiently large, the number bn is
smaller than g. Only a finite
number of members of the set can be larger than bn.
Thus, g cannot be a point of
accumulation, so that b is in
fact the largest number corresponding to a point
of accumulation. It is called the upper limit 
of
the set.
If we agree to choose in the construction the first interval
of the two (that with the smaller numbers as end-points) whenever
it contains an infinity of points of the set, we arrive in the
same way at the lower limit 
of the set.
The upper limit b and the lower limit a need not belong to the set. For example, in the case of the set of numbers an = 1/n, a2n-1 = (n - l)/n, we havea = 0, b = 1, but the numbers 0 and 1 are not members of the given set.
In the example, just given, the set contains no number larger than 1. We say that in this case 1, besides being the upper limit, is the upper bound G of the set with the general definition: The number G is called the upper bound of a set of numbers, if the set contains no number larger than G, and, if every number less than G is exceeded by at least one number belonging to the set.
It is important to note the distinction between the upper limit and the upper bound of a set. For example, consider the set of numbers 1, 1/2, 1/3, ··· . The upper bound is 1, the upper limit is 0, the number 0 giving the only point of accumulation of the set..
We shall now show that every set of numbers which is bounded above has an upper bound. A set of numbers is said to be bounded above if there is a number M such that all members of the set are smaller than M. First of all, note that if the set contains a largest member G, then 0 G is the upper bound of the set. But a set which is ebound above need not have a largest member, as is shown by the example (it — l)/n, n = l, 2, ···. We now assert that if the set has no largest member, its upper limit is also its upper bound.
In fact, let the set contain a number x > b. We consider all members of the set which are not less than x. There can only be a finite number of these, because otherwise the interval (x|M) would contain an infinity of members of the set and thus at least one point of accumulation, contrary to the assumption that b is the upper limit. Among the finite number of members of the set which are not less than x there would be a largest one and this member would be at the same time the largest member of the entire set. Thus, we should be returned to the case already dealt with. It follows that, if the set contains no largest member, then no member of the set exceeds the upper limit. The number b also fulfils the second condition that it should be the upper bound. In fact, let y be any number less than b; then the interval (y|M) is a neighbourhood of b. However, since b is a point of accumulation, its neighbourhood contains an infinity of points of the set, all larger than y.
The lower bound g of a set of numbers is correspondingly defined as that number which is not larger than any member of the set and which has the property that every number larger than g is also larger than at least one member of the set. Every set which is bounded below has a lower bound, which is either the smallest member of the set or else the lower limit of the set.
5. Convergent Sequences:. We consider now sequences of numbers a1, a2, ··· , always assuming that they are bounded. The principle of the point of accumulation shows that the set of numbers a1, a2, ··· has at least one point of accumulation. A sequence of numbers is called convergent if it has only one point of accumulation a. This number a is then called the limit of the sequence, and we write
The following definition is clearly equivalent to the one just given. A sequence of numbers a1, a2, ··· has the limit a , if and only if every neighbourhood of a contains all the members an of the sequence, with the possible exception of a finite number of members.
In fact, if the bounded sequence an has only one point of accumulation a, then only a finite number of members can lie outside any neighbourhood of a; otherwise, there would be some other point of accumulation. Conversely, if all neighbourhoods of a contain all the numbers an , with only a finite number of ex ceptions) then the sequence an is certainly bounded. It can only possess the one point of accumulation a. In fact. if a ' were another one, we could choose quite separate neighbourhoods of a and a', and in each of these there would be an infinity of numbers belonging to the sequence. This would contradict the assumption that only a finite number of members of the sequence lie outside any neighbourhood of a.
A sequence which does not have a limit should not be
regarded is anything abnormal. On the contrary, the existence of
a limit is in a sense exceptional. For example, the sequence with
the members a2n=1/2n, a2n-1=(n-1)/n,
n=1,2, ··· has two points of accumulation, namely 0 and
1.
The aggregate of the positive rational numbers can be regarded as a sequence of numbers, although we must first entirely dislocate the ordering by magnitude. The simplest way to arrive at such a sequence is to order the members by means of the array in Fig. 1. The line drawn in the figure shows the order in which the numbers should be taken, any number which has already appeared in the sequence being disregarded. As has already been mentioned, the set of all rational numbers has every real number as a point of accumulation.
The concept of convergence enables us to make a very useful deduction from the principle of the point of accumulation: If M is a given bounded, infinite set of numbers with x as point of accumulation, then M contains an infinite sequence a1, a2, ··· of numbers converging to the limit x.
In order to prove this, we assume that x is given by a nest of intervals c where an < x <bn. Since x is a point of accumulation, (a1 ,b1) contains an infinity of points of M. Choose one of these and call it a1 Again, (a2 ,b2) contains points of M. Choose one of these and call it a2, etc. The resulting sequence a1, a2, ··· is bounded and cannot have a point of accumulation other than x. Hence It converges to the limit x.We now call attention to two theorems on convergent sequences, which, though simple, are important in the sequel.If the sequence a1, a2, ···. converges to the limit a, then every infinite sub-sequence converges to a. For instance, a1,a2,a3,··· converges to aThis follows immediately from the observation that any point of accumulation of a sub-sequence must be a point of accumulation of the original sequence. An infinite sub-sequence must have at least one point of accumulation, and this can only be a. If a1, a2, ··· and b1, b2, ··· are sequences with the same limit g, then the mixed sequence a1, b1, a2, b2, ··· converges to g. Any neighbourhood of g contains all the numbersan and all the numbers bn, with the possible exception of a finite number of members of each sequence, whence it contains all members of the mixed sequence, except possibly for a finite number of these.
6.
Bounded Monotonic Sequences: A sequence of numbers a1,
a2, ··· is said to be monotonic
if either 
for all values of n
or 
for all values of n.
In the first case, we say that the sequence is monotonic non-decreasing
and in the second case that it is monotonic
non-increasing. We now prove the important
statement that every bounded monotonic
sequence is convergent. We may restrict ourselves
to the proof for the non-decreasing sequence. The other case is
exactly similar. Since every bounded sequence has at least one point of accumulation,
we need only show that our monotonic sequence cannot possess more
than one such point.Let there be two such points, a and
a', say, and a < a'. Construct around a
and a' two quite separate neighbourhoods Ua
and Ua' . Each must contain an infinity of members an
of the sequence. Take one of the members contained in Ua'
, say ar. Now let as
be the first member beyond ar which
lies in Ua. There must be such a
member, because Ua contains an
infinity of members. Now, all the members in Ua
are smaller than any in Ua', whence
ar>as
which contradicts the hypothesis that the sequence is non-decreasing. We may
add the following remark: If a1, a2,
··· is non-decreasing
and bounded, then 
for
every N. In fact, only a finite number of members an,
i.e., say a1, a2, ··· ,
aN-1 can be less than aN.
Hence, the limit is not less than aN.
In the same way, we see that the limit of a non-increasing
sequence is not larger than any member of the
sequence.
7. Cauchy's Convergence Test for
Sequences of Rational Numbers: Before we can
lay the foundations of calculation
with real numbers, we require a convergence test
which is not restricted to
sequences of rational numbers; but we cannot
formulate this until we have defined subtraction
for real numbers. Hence we shall prove here the
convergence test for rational numbers and return to the general
case in 9. p. 58B.The test
in question is: A sequence of rational
numbers a1, a2, ···
converges if, and only if, we can find corresponding to every
positive number e, however small, a number N(e) such that for every n > N
and m > N 
We shall first
show that if this inequality is satisfied for all sufficiently
large numbers m and n, then the sequence is
convergent. The boundedness of the sequence is proved as follows:
We take the special value e =
1. Then, for a sufficiently large value of n and all
sufficiently large values of m 
Attention must be drawn to the fact that the elements of the sequence a1, a2, ··· are assumed to he rational, but that this is not the case with the limit a.
Then, with a finite number of possible exceptions, all the numbers am lie in the interval ( an-1 |an+1). Thus, a properly chosen interval will contain all the numbers am without exception..The principle of the point of accumulation shows that the sequence has at least one point of accumulation. We must still show that there cannot be more than one. Suppose there are two such points, say, a and a'. We could construct about a and a' quite separate neighbourhoods (c|d) and (c'|d') so that
where we assume, as we may without restriction of generality, that a <a'. Since a and a' are assumed to be points of accumulation, (c|d) contains an infinite number of points an and (c'|d') an infinite number of points am. Thus, in particular, for an infinite set of values of n and m we have
But this contradicts the assumption, which shows that for all sufficiently large values of m and n
Hence, the sequence has one, and only one, point of accumulation.
We next show that, if the sequence a1, a2, ··· converges to a, then for every e > 0 and for all sufficiently large values of n and m
We take a neighbourhood (c|d) of a, the length (d - c) of which is less than or equal to e. If N is suitably chosen, then whenever n exceeds N, an lies in (c|d). Thus, if n > N and m > N, both am and an lie in (c|d), whence
8. Calculation with real numbers: So far, our work has given us the definition of real numbers by means of nests of intervals, and their ordering by magnitude. The theorem last proved provides simple means of defining the rules of arithmetical calculation with real numbers.
Let a real number a be given by a nest of intervals ( an| bn). Since the intervals form a nest, the numbers an form a monotonic non-decreasing sequence and the number bn a monotonic non-increasing sequence. These sequences are bounded; in fact, we may note that every an is less than or equal to b1, and every bn larger than or equal to a1, whence the sequences converge. In both cases, moreover, the limit is the real number a. In fact, every neighbourhood of a contains all the intervals ( an| bn), except possibly for a finite number of these, and thus the neighbourhood contains all but a finite number of members of the an and bn sequences. Hence, we may say that every real number can be exhibited as the limit of sequences of rational numbers.
If we now wish to define any operation of arithmetic for two real numbers a and b, we choose two sequences an and bn of rational numbers with the limits a and b, respectively. We perform the operation on the pairs of numbers an and bn and thus obtain a new sequence. When we have proved that this sequence has a limit, we shall say, by way of definition, that it is the result of the operation on the two real numbers a and b.
Let a and b be two arbitrary real numbers and
We consider the sequences
If we can prove that these sequences converge, we can set up the definitions
The convergence of these sequences will be proved by means of Cauchy's convergence test.
It follows from the convergence of a1, a2, ··· that, if e is a given positive number and n and m are sufficiently large, say n > N1, and m > N1, then
and, from the convergence of b1, b2, ···, that if n and m are sufficiently large, say n > N2 and m > c, then
If N(e) denotes the larger of the two numbers N1 and N2, then, if n > N(e), m > N(e),
and
By Cauchy's test, both the sequences an + bn and an - bn converge.
In order to prove that anbn converges, we must first note that the numbers an + bn form bounded sets. There are therefore two positive rational numbers A and B such that for all values of n
Now,
Since the sequences a1, a2, ··· and b1, b2, ··· converge, we can find numbers N1 and N2 corresponding to any given e > 0, such that
and
Thus, if n and m are both larger than the larger one of the two numbers N1 and N2, the above inequalities hold simultaneously, whence we have
Cauchy's test shows at once that the sequence anbn converges.
We now suppose that a ¹
0 and 
We must show that 1/an
converges. It is first necessary to show that if n is
sufficiently large, |an| is larger
than a positive number p independent of n. We
take a rational neighbourhood of a which does not
contain 0. This is possible, since a ¹
0. From a suitable n = n0
onwards, all the members of the sequence a1,
a2, ··· lie in this neighbourhood. This
shows that, for n > n0,
|an| ³ p, where p is
the absolute value corresponding to the end-point of the interval
that is nearer to 0. The convergence of the sequence 1/a1, 1/a2,
··· is not affected by the omission of the first n0 members,whence we may now assume
that for all values of n
We observe that
Let e >0 be given. If N is suitably chosen,.then, since a1, a2, ··· converges, n > N and m > N give
This proves the convergence of 1/an, provided that a ¹ 0. Obviously, any real number may be exhibited as the limit of more than one sequence of rational numbers. It might be thought that the definitions given above do not define the arithmetical operations uniquely. For instance, let
give one representation of the numbers a and b and
another. Then possibly the two sequences an + bn and an' + bn' might have different limits. (We have proved already that they do have limits.) We shall now prove that this difficulty does not arise. It will be shown that if
then
and, if
The .proof is very simple. We have already shown that, if
then the mixed sequence a1, a1', a2, a2', ··· has the limit a. In the same way, we see that b1, b1', b2, b2', ··· converges
We find from this and the above theorems that the mixed sequences a1 ± b1, a2 ± b2, ··· and a1b1, a2'b2', ··· and, if a ¹ 0, 1/a1,1/a1', ··· converge.
It has already been proved that every sub-sequence of a given convergent sequence converges to the same limit. It follows from this that the sequences
which are sub-sequences of a convergent sequence, must converge to the same limit. In the same way,
have the same limit, and the same is true for
These results allow us to settle another important question which is connected with our definitions of the operations of arithmetic.
The class of real numbers contains the rational numbers. In the course of our definitions of operations on the real numbers, we have thus incidentally defined these operations for the rational numbers. But we began by assuming that operations on rational numbers are known. We must therefore verify that the new definitions do not give rise to any contradiction in the case of rational numbers. What we must show is that if
are rational numbers, then
and, if a ¹ 0,
We should first note that a rational number a is the limit of the rational sequence a, a, ···. For the two sequences a1, a1', ··· and b1, b1', ···, we may take the special sequences a, a, ··· and b, b, ···. The above theorems then yield
which is the required result.
It need hardly be mentioned that, as a result of our definitions, all the rules of calculation which hold for rational numbers also hold for all real numbers. We need only apply the rules to the rational numbers forming the sequences. Let us, for instance, prove the distributive law
Let
Then the left-hand side of the equality to be proved is
and the right hand side is
But since the distributive law is true for the rational numbers, the two sequences are the same, and this must also be true of their limits.
9. The General Form of Cauchy's Convergence Test:. We return to Cauchy's convergence test, which we have already proved for rational sequences. Now, that the operations of arithmetic, in particular, subtraction have been established for real numbers, we can formulate the convergence test quite generally for real numbers. The sequence a1, a2, ··· converges if, and only if, for any given e > 0, we can find a subscript N(e) such that whenever m and n are both larger than N(e)
The proof is exactly like that given earlier, and need not be repeated.
The following fact is of great theoretical importance. Cauchy's convergence test contains in its formulation a means of estimating errors. In fact, if we are given the sequence and know the number N(e), we can state at once that the limit of the sequence lies between the numbers an + e and an - e whenever n > N(e).
In this respect, Cauchy's test differs from the test for monotonic sequences. The latter proves the existence of the limit, but gives no means of estimating the limit. Thus, in proofs of convergence which depend on this test any estimation of the limit (and theoretically it is always necessary to give one!) must depend on. separate and extraneous considerations.