Examples of convergent sequences

1. Introduction

Recall that a sequence ( a n ) converges to l if

ε > 0   N   n   ( n N | a n - l | < ε )

This page gives three examples of convergent sequences, all properly proved. The exercises for this page give further examples for you to look at.

2. The constant sequence

Consider first the constant sequence where each term is a n = l .

Proposition.

( a n ) converges to l .

Proof.

We must prove that ε > 0   N   n   ( n N | a n - l | < ε ) .

Remark.

For this particular proof, the statement starts with a so we must start by letting ε > 0 be arbitrary. Then we must give a value of N for the . In fact it won't matter what N we take, so we take N = 1 . The rest of the proof is done by letting n N be arbitrary, and is easy.

Subproof.

Let ε > 0 in be arbitrary.

Subproof.

Let N be 1 .

Subproof.

Let n be arbitrary.

Subproof.

Assume n N .

Then a n = l so | a n - l | < ε since ε > 0 .

Hence ( n N | a n - l | < ε ) .

Hence n   ( n N | a n - l | < ε ) .

Hence N   n   ( n N | a n - l | < ε ) .

Hence ε > 0   N   n   ( n N | a n - l | < ε ) .

3. The sequence l + 1 / n

Now consider the sequence ( b n ) with terms b n = l + 1 / n .

Proposition.

( b n ) converges to l .

Proof.

Remark.

Again we must start by letting ε > 0 be arbitrary. Then we must give a value of N for the . Because we have already chosen ε by this stage it is OK to choose N in a way depending on ε , and this is what we will do, using the x integer-part function.

Subproof.

Let ε > 0 in be arbitrary.

Subproof.

Let N be 1 + 1 / ε .

So 1 + 1 / ε N , so 1 / ε N - 1 , hence 1 / ε < N , hence ε > 1 / N since these terms are all positive.

Subproof.

Let n be arbitrary.

Subproof.

Assume n N .

Then | b n - l | = 1 / n .

So | b n - l | = 1 / n 1 / N < ε .

Hence ( n N | b n - l | < ε ) .

Hence n   ( n N | b n - l | < ε ) .

Hence N   n   ( n N | b n - l | < ε ) .

Hence ε > 0   N   n   ( n N | b n - l | < ε ) .

4. The sequence x n

Let x > 0 be real. Define a sequence ( c n ) by c n = x n . To analyse this sequence the following result is helpful.

Bernoulli's inequality.

Let x > -1 and n . Then ( 1 + x ) n 1 + n x .

Proof.

By induction on n . For n = 0 the inequality is just 1 1 , which is true. If n 0 and ( 1 + x ) n 1 + n x holds, then as 1 + x > 0 , we have ( 1 + x ) n + 1 1 + n x 1 + x = 1 + ( n + 1 ) x + n x 2 1 + ( n + 1 ) x , and this proves the induction step.

Theorem.

If 0 < x < 1 and c n = x n then c n 0 as n .

Proof.

Remark.

We have to estimate x n . Write x = 1 1 + y where y > 0 , and observe that this means x n = 1 ( 1 + y ) n 1 1 + n y < 1 n y , and this gives the main idea of the proof.

Subproof.

Let ε > 0 be arbitrary.

Let y = 1 x - 1 , and note that y > 0 since 0 < x < 1 .

Let N be greater than 1 ε y , such as N = 1 ε y + 1 .

Subproof.

Let n N be arbitrary.

Then | c n | = x n < 1 n y and 1 n y 1 N y < ε since n N and N y > y ε y = 1 ε .

Thus | c n | < ε .

So n   ( n N | c n | < ε ) , as n N was arbitrary.

So N   n   ( n N | c n | < ε ) .

So ε > 0   N   n   ( n N | c n | < ε ) .

The cases for other values of x are proved similarly.

The cases of other values of x are covered in the exercises. For x = 0 or 1 the sequence c n = x n is a constant sequence and hence convergent. For x = -1 , c n = x n is nonconvergent and for x > 1 and x < -1 , c n = x n is an unbounded and hence nonconvergent sequence.

5. The sequence n + 1 - n

Now consider the sequence ( d n ) with terms d n = n + 1 - n .

Proposition.

( d n ) converges to 0 .

Proof.

Remark.

We must start by letting ε > 0 be arbitrary. Then we must give a value of N for the . A little rough calculation before we start helps here.

n + 1 - n = n + 1 - n n + 1 + n n + 1 + n = 1 n + 1 + n

Now we are ready to give the proof.

Subproof.

Let ε > 0 in be arbitrary.

Subproof.

Let N be 1 4 ε 2 .

So N 1 4 ε 2 , hence 2 N 1 ε , hence 1 2 N ε since these terms are all positive.

Subproof.

Let n be arbitrary.

Subproof.

Assume n N .

Then | d n | = 1 n + n + 1 .

So | d n | < 1 2 n 1 2 N ε .

So | d n | < ε .

Hence ( n N | d n | < ε ) .

Hence n   ( n N | d n | < ε ) .

Hence N   n   ( n N | d n | < ε ) .

Hence ε > 0   N   n   ( n N | d n | < ε ) .

Proofs like this should not be so difficult. There is always an algebraic calculation to make, and it might help to do this in rough first. The proof itself follows the format of the proof rules exactly, and once the rough calculation has been done can be written out almost without any thinking at all.

Exercise.

Show, using similar methods, that the sequence defined by e n = n 2 + 2 n - n converges to 1 .