# Solutions to “next number in sequence”: preRMO, pRMO and RMO

What is the next number in sequence?

A) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

Solution to A:

Ans is 20. The sequence is the position in the letter of the alphabet of the first letter in the numbers 1 to 12, when given in full. e.g. ONE: O=15.

B) 1, 8, 11, 18, 80, ?

Ans is 81. The sequence comprises whole numbers beginning with a vowel.

C) 1, 2, 4, 14, 21, 22, 24, 31, ?

Ans is 32, The sequence comprises whole numbers containing the letter O.

D) 4, 1, 3, 1, 2, 4, 3, ?

Ans. is 2. The sequence is as follows: there is one number between the two I’s, two numbers between the two 2’s, three numbers between the two 3’s and four numbers between the two 4’s.

E) 1, 2, 4, 7, 28, 33, 198, ?

Ans is 205. $1 + 1 \times 2 + 3 \times 4 + 5 \times 6 + 7$

F) 17, 8, 16, 23, 28, 38, 49, 62, ?

Answer is 70. Sum of digits in all previous numbers in the sequence.

G) 27, 216, 279, 300, ?

Ans is 307. Difference divided by 3 and added to the last number.

H) 9,7,17,79,545, ?

Answer is 4895. Each number is multiplied by its rank in the sequence, and the next number is subtracted.

$9 \times 1 - 2 = 7 \times 3 -4 = 17 \times 5 - 6 = 79 \times 7 - 8 = 545 \times 9 - 10 = 4895$

I) 2,3,10,12,13, 20,?

Answer is 21. They all begin with the letter T.

J) 34, 58, 56, 60, 42, ?

Answer is 52. The numbers are the totals of the letters in the words ONE, TWO, THREE, FOUR, FIVE, SIX when A=1, B=2, C=3, etc.

Regards,

Nalin Pithwa.

# Next number in sequence: PreRMO, pRMO, RMO

What is the next number in the sequence?

a) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

b) 1,8,11, 18, 80, ?

c) 1,2,4,14,21,22,24,31,?

d) 4,1,3,1,2,4,3,?

e) 1,2,4,7,28,33,198,?

f) 17,8,16,23, 28, 38, 49, 62, ?

g) 27, 216, 279, 300, ?

h) 9,7,17,79,545,?

i) 2,3,10,12,13,20,?

j) 34, 58, 56, 60, 42, ?

Regards,

Nalin Pithwa.

# Set Theory, Relations, Functions: Preliminaries: Part VIII

SETS and FUNCTIONS:

Reference: Introductory Real Analysis by A. N. Kolmogorov and S V Fomin, Dover books.

Operations on sets:

Let A and B be any two sets. Then, by the sum or union of A and B, denoted by $A \bigcup B$, is meant the set consisting of all elements which belong to at least one of the sets A and B. More generally, by the sum or union of an arbitrary number (finite or infinite) of sets $A_{\alpha}$ (indexed by some parameter $\alpha$), we mean the set, denoted by $\bigcup_{\alpha}A_{\alpha}$ of all elements belonging to at least one of the sets $A_{\alpha}$.

By the intersection $A \bigcap B$ of two given sets A and B, we mean the set consisting of all elements which belong to both A and B. For example, the intersection of the set of all even numbers and the set of all integers divisible by 3 is the set of all integers divisible by 6. By the intersection of an arbitrary number (finite or infinite) of sets $A_{\alpha}$, we mean the set, denoted by $\bigcap_{\alpha}A_{\alpha}$ of all elements belonging to every one of the sets $A_{\alpha}$. Two sets A and B are said to be disjoint if $A \bigcap B=\phi$, that is, if they have no elements in common. More generally, let $\mathscr{F}$ be a family of sets such that $A \bigcap B=\phi$ for every pair of sets A, B in $\mathscr{F}$. Then, the sets in $\mathscr{F}$ are said to be pairwise disjoint.

It is an immediate consequence of the above definitions that the operations $\bigcup$ and $\bigcap$ are commutative and associative, that is,

$A \bigcup B=B \bigcup A$ and $(A \bigcup B) \bigcup C= A \bigcup (B \bigcup C)$; $A \bigcap B=B \bigcap A$; and, $(A \bigcap B)\bigcap C = A \bigcap (B \bigcap C)$.

Moreover, the operations $\bigcup$ and $\bigcap$ obey the following distributive laws:

$(A \bigcup B)\bigcap C = (A \bigcap C) \bigcup (B \bigcap C)$….call this I

$(A \bigcap B)\bigcup C = (A \bigcup C) \bigcap (B \bigcup C)$….call this II.

For example, suppose that $x \in (A \bigcup B)\bigcap C$, so that x belongs to the left-hand side of I. Then, x belongs to both C and $A \bigcup B$, that is, x belongs to both C and at least one of the sets A and B. But then x belongs to at least one of the sets $A \bigcap C$ and $B \bigcap C$, that is, $x \in (A \bigcap C)\bigcup (B \bigcap C)$, so that x belongs to the right hand side of I. Conversely, suppose that $x \in (A \bigcap C)\bigcup (B \bigcap C)$. Then, x belongs to at least one of the two sets $A \bigcap C$ and $B \bigcap C$. It follows that x belongs to both C and at least one of the two sets A and B, that is, $x \in C$ and $x \in A \bigcup B$, or equivalently $x \in (A \bigcup B)\bigcap C$. This proves I, and II is proved similarly.

By the difference of two sets, $A - B$ between two sets A and B (in that order), we mean the set of all elements of A which do not belong to B. Note that it is not assumed that $A \supset B$. It is sometimes convenient (e.g., in measure theory) to consider the symmetric difference of two sets A and B, denoted by $A \delta B$ and defined as the union of the two differences $A-B$ and $B-A$: $A \delta B=(A-B)\bigcup (B-A)$.

We will often be concerned later with various sets which are all subsets of some underlying basic set R, for example, various sets of points on the real line. In this case, given a set A, the difference $R-A$ is called the complement of A, denoted by $A^{'}$.

An important role is played in set theory and its applications by the following duality principle:

$R - \bigcup_{\alpha}A_{\alpha}=\bigcap_{\alpha}(R-A_{\alpha})$…call this III

$R - \bigcap_{\alpha}A_{\alpha}=\bigcup_{\alpha}(R-A_{\alpha})$…call this IV.

In words, the complement of a union equal the intersection of the complements; and, the complement of an intersection equals the union of the complements. According to the duality principle, any theorem involving a family of subsets of a fixed set R can be converted automatically into another, “dual” theorem by replacing all subsets by their complements, all unions by intersections and all intersections by unions. To prove III, suppose that $x \in R - \bigcup_{\alpha}A_{\alpha}$….call this V.

Then, x does not belong to the union $\bigcup_{\alpha}A_{\alpha}$. …call this VI.

That is, x does not belong to any of the sets $A_{\alpha}$. It follows that x belongs to each of the complements $R - \bigcup_{\alpha}A_{\alpha}$, and hence, $x \bigcap_{\alpha}(R-A_{\alpha})$….call this VII.

Conversely, suppose that VII holds, so that x belongs to every set $R - A_{\alpha}$. Then, x does not belong to any of the sets $A_{\alpha}$, that is, x does not belong to the union VI, or equivalently V holds true. This proves 3.

Homework: Prove IV similarly.

Remark:

The designation “symmetric difference” for the set $A \Delta B$ is not too apt, since $A \Delta B$ has much in common with the sum $A \bigcup B$. In fact, in $A \bigcup B$ the two statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in $A \Delta B$ the same two statements are joined by “or” used in the ordinary “either…or…” sense (as in “to be or not to be”). In other words, x belongs to $A \bigcup B$ if and only if x belongs to either A or B or both, while x belongs to $A \Delta B$ if and only if x belongs to either A or B but not both. The set $A \Delta B$ can be regarded as a kind of “modulo-two sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B).

Functions and mappings. Images and preimages:

(Of course, we have dealt with this topic in quite detail so far in the earlier blog series; but as presented by Kolmgorov and Fomin here, the flavour is different; at any rate, it helps to revise. Revision refines the intellect.) 🙂 🙂 🙂

A rule associating a unique real number $y=f(x)$ with each element of a set real numbers X is said to define a (real) function f on X. The set X is called the domain (of definition) of f, and the set Y of all numbers $f(x)$ such that $x \in X$ is called the range of f.

More generally, let M and N be two arbitrary sets. Then a rule associating a unique element $b=f(a) \in N$ with each element $a \in M$ is again said to define a function f on M (or a function f with domain M). In this more general context, f is usually called a mapping of f M into N. By the same token, f is said to map M into N(and a into b).

If a is an element of M, the corresponding element $b=f(a)$ is called the image of a (under the mapping f). Every element of M with a given element $b \in N$ as its image is called a preimage of b. Note that in general b may have several pre-images. Moreover, N may contain elements with no pre-images at all. If b has a unique pre-image, we denote this pre-image by $f^{-1}(b)$.

If A is a subset of M, the set of all elements $f(a) \in N$ such that $a \in A$ is called the image of A, denoted by $f(A)$. The set of all elements of M whose images belong to a given set $B \subset N$ is called the preimage of B, denoted by $f^{-1}(B)$. If no element of B has a preimage, then $f^{-1}(B)=\phi$. A function f is said to map M into N if $f(M)=N$, as is always the case, and onto N if $f(M)=N$. Thus, every “onto” mapping is an “into” mapping, but converse is not true.

Suppose f maps M onto N. Then, f is said to be one-to-one if each element $b \in N$ has a unique preimage $f^{-1}(b)$. In this case, f is said to establish a one-to-one correspondence between M and N, and the mapping $f^{-1}$ associating $f^{-1}(b)$ is called the inverse of f.

Theorem I:

The preimage of the union of two sets is the union of the preimages of the sets $f^{-1}(A \bigcup B)=f^{-1}(A) \bigcup f^{-1}(B)$.

Proof of Theorem I:

If $x \in f^{-1}(A \bigcup B)$, then $f(x) \in A \bigcup B$ so that $f(x)$ belongs to at least one of the sets A and B. But, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcup f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcup f^{-1}(B)$, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$. Therefore, $f(x)$ belongs to at least one of the sets A and B, that is, $f(x) \in A \bigcup B$. But, then $x \in f^{-1}(A \bigcup B)$.

QED.

Theorem 2:

The preimage of the intersection of two sets is the intersection of the preimages of the sets:

$f^{-1}(A \bigcap B)=f^{-1}(A) \bigcap f^{-1}(B)$.

Proof of Theorem 2:

If $x \in f^{-1}(A \bigcap B)$, then $f(x) \in A \bigcap B$, so that $f(x) \in A$ and (meaning, simultaneously) $f(x) \in B$. But, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcap f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcap f^{-1}(B)$, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$. Therefore, $f(x) \in A$ and $f(x) \in B$, that is, $f(x) \in A \bigcap B$. But, then $x \in f^{-1}(A \bigcap B)$.

QED.

Theorem 3:

The image of the union of two sets equals the union of the images of the sets $f(A \bigcup B) = f(A) \bigcup f(B)$.

Proof of theorem 3:

If $y \in f(A \bigcup B)$, then $y=f(x)$ where x belongs to at least one of the sets A and B. Therefore, $y=f(x)$ belongs to at least one of the sets $f(A)$ and $f(B)$. That is, $x \in f(A) \bigcup f(B)$.

Conversely, if $y \in f(A) \bigcup f(B)$, then $y=f(x)$. where x belongs to at least one of the sets A and B, that is, $x \in A \bigcup B$ and hence, $y=f(x) \in f(A \bigcup B)$.

QED.

Remark I:

Surprisingly enough, the image of the intersection of two sets is not so “well-behaved”. The image of the intersection of two sets does not necessarily equal the intersection of the images of the sets. For example, suppose the mapping f projects the xy-plane onto the x-axis, carrying the point $(x,y)$ into the $(x,0)$. Then, the segments $0 \leq x \leq 1$ with $y=0$, and $0 \leq x \leq 1$ with $y=1$ do not intersect, although their images coincide.

Remark 2:

In the light of above remark: consider the following: If a function is not specified on elements, it is important in general to check that f is well-defined. That is, unambiguously defined. For example, if the set A is the union of two subsets $A_{1}$ and $A_{2}$, and we are considering a function $f: A \longrightarrow B$, then one can try to specify a function from set A to the set B $\{ 0,1\}$ by declaring that f is to map everything in $A_{1}$ to 0 and is to map everything in $A_{2}$ to 1. This unambiguously defines f unless $A_{1}$ and $A_{2}$ have elements in common in which case it is not clear whether these elements should map to 0 or to 1. Checking that this f is well-defined therefore amounts to checking that $A_{1}$ and $A_{2}$ have no intersection.

Remark 3:

Theorems 1-3 continue to hold for unions and intersections of an arbitrary number (finite or infinite) of sets $A_{\alpha}$:

$f^{-1}(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f^{-1}(A_{\alpha})$

$f^{-1}(\bigcap_{\alpha}A_{\alpha})=\bigcap_{\alpha}f^{-1}(A_{\alpha})$

$f(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f(A_{\alpha})$

Decomposition of a set into classes. Equivalence relations.

Decompositions of a given set into pairwise disjoint subsets play an important role in a great variety of problems. For example, the plane (regarded as a point set) can be decomposed into lines parallel to the x-axis, three dimensional space can be decomposed into concentric spheres, the inhabitants of a given city can be decomposed into different age groups, and so on. Any such representation of a given set M as the union of a family of pairwise disjoint subsets of M is called a decomposition or partition of M into classes.

A decomposition is usually made on the basis of a certain criterion, allowing us to assign the elements of M to one class or another. For example, the set of all triangles in the plane can be decomposed into classes of congruent triangles, or classes of triangles of equal area, the set of all functions of x can be decomposed into classes of functions all taking the same value at a given point x, and so on. Despite the great variety of such criteria, they are not completely arbitrary. For example, it is obviously impossible to partition all real numbers into classes by assigning the number b to the same class as number a if and only if $b>a$. In fact, if $b>a$, b must be assigned to same class as a but then a cannot be assigned to same class as b, since $a. Moreover, since a is not greater than itself, a cannot be assigned to the class containing itself!! As another example, it is impossible to partition the points of the plane into classes by assigning two points to the same class if and only if the distance between them is less than 1. In fact, if the distance between a and b is less than 1, and if the distance between b and c is less than 1, it does not follow that distance between a and c is less than 1. (Hint: Think of triangle inequality). Thus, by assigning a to the same class as b and b to the same class as c, we may well find that two points fall in the same class even though the distance between them is greater than 1!

These examples suggest conditions that which must be satisfied by any criterion it it is to be used as the basis for partitioning a given set into classes. Let M be a set, and let certain ordered pairs (a,b) of elements of M be called “labelled.” If $(a,b)$ is a labelled pair, we say that a is related to b by the binary relation R and we write it symbolically as aRb. For example, if a and b are real numbers, $aRb$ might mean $a, while if a and b are triangles, $aRb$ might mean that a and b have the same area. A relation between elements of M is called a relation on M, if there exists at least one labelled pair $(a,b)$ for every $a \in M$. A relation R on M is called an equivalence relation (on M) if it satisfies the following three conditions:

• Reflexivity: $aRa$ for every $a \in M$.
• Symmetry: If $aRb$, then $bRa$.
• Transititivity: If $aRb$ and $bRc$, then $aRc$.

🙂 🙂 🙂

Nalin Pithwa, more later

# Set theory, relations, functions: preliminaries: Part VII

FINITE, COUNTABLE AND UNCOUNTABLE SETS:

(Reference: Principles of Mathematical Analysis, Third Edition, Walter Rudin:)

We begin with the function concept.

Definition: Consider two sets A and B whose elements may be any objects whatsoever, and suppose that with each element x of A there is associated in some manner, an element of B, which we denote by $f(x)$. Then, f is said to be a function from A to B (or a mapping of A into B). The set A is called the domain of f (we also say f is defined on A), and the elements f(x) are called the values of f. The set of all values of f is called the range of f.

Definition: Let A and B be two sets and let f be a mapping of A into B. If $E \subset A$, then $f(E)$ is defined to be the set of all elements f(x), for $x \in E$. We call $f(E)$ the image of E under f. In this notation, $f(A)$ is the range of f. It is clear that $f(A) \subset B$. If $f(A)=B$, we say that f maps A onto B. (Note that, according to this usage, onto is more specific than into).

If $E \subset B$, $f^{-1}(E)$ denotes the set of all $x \in A$ such that $f(x) \in E$. We call $f^{-1}(E)$ the inverse image of E under f. If $y \in B$, then $f^{-1}(y)$ is the set of all $latex \in A$ such that $f(x)=y$. If, for each $y \in B$, $f^{-1}(y)$ consists of at most one element of A, then f is said to be a 1-1 (one-one) mapping of A into B. This may also be expressed as follows: f is a 1-1 mapping of A into B provided that $f(x_{1}) \neq f(x_{2})$ whenever $x_{1} \neq x_{2}$ and $x_{1} \in A$ and $x_{2} \in A$.

(The notation $x_{1} \neq x_{2}$ , means that $x_{1}$ and $x_{2}$ are distinct elements, otherwise we write $x_{1} = x_{2}$).

Definition:

If there exists a 1-1 mapping A of onto B, we say that A and B can be put in 1-1 correspondence, or that A and B have the same cardinal number, or briefly, that A and B are equivalent, and we write $A \sim B$. This relation clearly has the following properties:

It is reflexive: $A \sim A$

It is symmetric: If $A \sim B$, then $B \sim A$.

It is transitive: If $A \sim B$, and $B \sim C$, then $A \sim C$.

Any relation with these three properties is called an equivalence relation.

Definition:

For any positive integer n, let $J_{n}$ be the set whose elements are the integers $1, 2, \ldots, n$. And, let J be the set consisting of all positive integers. For any set A, we say:

(a) A is finite if $A \sim J_{n}$ for some n (the empty set is also considered to be finite).

(b) A is infinite, if A is not finite.

(c) A is countable if $A \sim J$.

(d) A is uncountable if A is neither finite nor countable.

(e) A is at most countable if A is finite or countable.

Countable sets are sometimes called enumerable, or denumerable.

For two finite sets A and B, we evidently have $A \sim B$ if and only if A and B contain the same number of elements. For infinite sets, however, the idea of “having the same number of elements” becomes quite vague, whereas the notion of 1-1 correspondence retains its clarity.

Example:

Let A be the set of all integers. Then, A is countable. For, consider the following arrangement of the sets A and J:

A: 0,1,-1,2,-2,3,-3,…

J:1,2,3,4,5,6,7,…

We can, in this example, even give an explicit formula for a function f from J to A which sets up a 1-1 correspondence:

When n is even, $f(n)=\frac{n}{2}$

When n is odd, $f(n)=-\frac{n-1}{2}$

Remark:

A finite set cannot be equivalent to one of its proper subsets. That this is, however, possible for infinite sets, is shown by the above example, in which J is a proper subset of A.

In fact, we can define infinite sets as follows: A set A is infinite if A is equivalent to one of its proper subsets.

Definition: SEQUENCES:

By a sequence, we mean a function f defined on the set J of all positive integers. If $f(n)=x_{n}$, for $n \in J_{n}$, it is customary to denote the sequence f by the symbol $\{ x_{n}\}$, or sometimes by $x_{1}, x_{2}, x_{3}, \ldots$. The values of f, that is, the elements $x_{n}$ are called the terms of the sequence. If A is a set and if $x_{n} \in A$, for all $n \in J$, then $\{ x_{n}\}$ is said to be a sequence in A, or a sequence of elements of A.

Note that the terms $x_{1}, x_{2}, x_{3}, \ldots$ of a sequence need not be distinct.

Since every countable set is the range of a 1-1 function defined on J, we may regard every countable set as the range of a sequence of distinct terms. Speaking more loosely, we may say that the elements of any countable set can be “arranged” in a sequence.

Sometimes, it is convenient to replace J in this definition by the set of all non-negative integers, that is, to start with 0 rather than with 1.

Regards,

Nalin Pithwa

PS: the above exposition will be extremely useful to RMO, INMO and IITJEE advanced maths students and students preparing for Chennai Mathematics Institute Entrance Examination.

# Intel Pentium P5 floating point unit error (1994) — an RMO problem !!!

Problem:

Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 litres of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that $p+2$ is also a prime number. Then, the first number theorist would pipette out 1/p litres of chemical and the second $\frac{1}{(p+2)}$ litres. How many times do they have to play this game to empty the flask completely?

Hint:

A bit of real analysis is required.

Reference:

I will publish the reference when I post the solution. So that all students/readers can curb their impulse to see the solution immediately!!!

I hope you will be hooked to the problem in a second….!!! Here is a beautiful utility of pure math! 🙂

Cheers,

Nalin Pithwa

PS: I do not know if the above problem did (or, will?? )appear as RMO question. It is just my wild fun with math to kindle the intellect of students in analysis !! 🙂

# Huygens’ problem to Leibnitz: solution

In the Feb 23 2018 blog problem, we posed the following question:

Sum the following infinite series:

$1+\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15}+ \ldots$.

Solution:

The sum can be written as:

$S=\sum_{n=1}^{\infty}P_{n}$, where $P_{n}=\frac{2}{n(n+1)}=2(\frac{1}{n}-\frac{1}{n+1})$.

Thus, $2(1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots)=2$. This is the answer.

If you think deeper, this needs some discussion about rearrangements of infinite series also. For the time, we consider it outside our scope.

Cheers,

Nalin Pithwa.

# Huygens’ Problem to Leibnitz

Young philosopher Leibnitz went to Huygens for training in mathematics. Huygens’ asked Leibnitz to find the sum of the following infinite series:

Question: $1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15}+ \ldots$ Calculate the sum.

Kindly share your ideas, even partial solutions are welcome.

Nalin Pithwa.

# Real Numbers, Sequences and Series: part 9

Definition.

We call a sequence $(a_{n})_{n=1}^{\infty}$ a Cauchy sequence if for all $\varepsilon >0$ there exists an $n_{0}$ such that $|a_{m}-a_{n}|<\varepsilon$ for all m, $n > n_{0}$.

Theorem:

Every Cauchy sequence is a bounded sequence and is convergent.

Proof.

By definition, for all $\varepsilon >0$ there is an $n_{0}$ such that

$|a_{m}-a_{n}|<\varepsilon$ for all m, $n>n_{0}$.

So, in particular, $|a_{n_{0}}-a_{n}|<\varepsilon$ for all $n > n_{0}$, that is,

$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.

Let $M=\max \{ a_{1}, \ldots, a_{n_{0}}, a_{n_{0}+1}+\varepsilon\}$ and $m=\min \{ a_{1}, \ldots, a_{n_{0}+1}-\varepsilon\}$.

It is clear that $m \leq a_{n} \leq M$, for all $n \geq 1$.

We now prove that such a sequence is convergent. Let $\overline {\lim} a_{n}=L$ and $\underline{\lim}a_{n}=l$. Since any Cauchy sequence is bounded,

$-\infty < l \leq L < \infty$.

But since $(a_{n})_{n=1}^{\infty}$ is Cauchy, for every $\varepsilon >0$ there is an $n_{0}=n_{0}(\varepsilon)$ such that

$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.

which implies that $a_{n_{0}+1}-\varepsilon \leq \underline{\lim}a_{n} =l \leq \overline{\lim}a_{n}=L \leq a_{n_{0}+1}+\varepsilon$. Thus, $L-l \leq 2\varepsilon$ for all $\varepsilon>0$. This is possible only if $L=l$.

QED.

Thus, we have established that the Cauchy criterion is both a necessary and sufficient criterion of convergence of a sequence. We state a few more results without proofs (exercises).

Theorem:

For sequences $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$.

(i) If $l \leq a_{n} \leq b_{n}$ and $\lim_{n \rightarrow \infty}b_{n}=l$, then $(a_{n})_{n=1}^{\infty}$ too is convergent and $\lim_{n \rightarrow \infty}a_{n}=l$.

(ii) If $a_{n} \leq b_{n}$, then $\overline{\lim}a_{n} \leq \overline{\lim}b_{n}$, $\underline{\lim}a_{n} \leq \underline{\lim}b_{n}$.

(iii) $\underline{\lim}(a_{n}+b_{n}) \geq \underline{\lim}a_{n}+\underline{\lim}b_{n}$

(iv) $\overline{\lim}(a_{n}+b_{n}) \leq \overline{\lim}{a_{n}}+ \overline{\lim}{b_{n}}$

(v) If $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$ are both convergent, then $(a_{n}+b_{n})_{n=1}^{\infty}$, $(a_{n}-b_{n})_{n=1}^{\infty}$, and $(a_{n}b_{n})_{n=1}^{\infty}$ are convergent and we have $\lim(a_{n} \pm b_{n})=\lim{(a_{n} \pm b_{n})}=\lim{a_{n}} \pm \lim{b_{n}}$, and $\lim{a_{n}b_{n}}=\lim {a_{n}}\leq \lim {b_{n}}$.

(vi) If $(a_{n})_{n=1}^{\infty}$, $(b_{n})_{n=1}^{\infty}$ are convergent and $\lim_{n \rightarrow \infty}b_{n}=l \neq 0$, then $(\frac{a_{n}}{b_{n}})_{n=1}^{\infty}$ is convergent and $\lim_{n \rightarrow \frac{a_{n}}{b_{n}}}= \frac{\lim {a_{n}}}{\lim{b_{n}}}$.

Reference: Understanding Mathematics by Sinha, Karandikar et al. I have used this reference for all the previous articles on series and sequences.

More later,

Nalin Pithwa