Sequences and Series.

Part I Sequences.

Definition 1. Let P be the set of positive integers. A sequence is simply a function from P into the set of real numbers R.

P is of course the set {1,2, ¼ }. Thus a function a: P ® R is a sequence.

The image a(n) is called the n-th term of the sequence and is also written as a _n ,

We also write (a₁ , a₂, ¼ ) or simply (a_n) for the sequence.

Here we use the round bracket for sequences. One should not confused the sequence (a₁ , a₂, ¼ ) with a row vector.

We are interested in the behaviour of the values or points of the sequences. We want to know if they are bunched together like a cluster or they become further and further apart or oscillatory. We focus on whether the points are bunched together or not. We have a technical term of this bunching together.

Definition 2. Let (a_n ) be a sequence in R. We say (a_n ) tends to a real number a in R if for any e > 0, there exists a positive integer N₀ such that for all n in P with n ³ N₀ , |a_n - a| < e.

That is,

n ³ N₀ Þ |a_n - a| < e

Notation:

If (a_n ) tends to a, we write

a_n ® a as n® ¥

or

Or just simply, a_n ® a .

Definition 3. We say (a_n ) converges if there exits a real number a such that a_n ® a, otherwise (a_n ) diverges or is divergent.

Example 4.

1. a_n = c for all n in P. This is a constant sequence obviously a_n ® c .

Given any e > 0 , take any positive integer N obviously for any n ³ N

|a_n - c| = |c - c| = 0 < e.

2. a_n = (-1) ⁿ . Then (a_n ) is divergent. There is a quick way to see this. Observe that the value changes from 1 to -1 and so there is no way it can get close to any value.

If you like the following is a proof of this fact.

For any a in R, by the triangle inequality,

|1- a| + |(-1)-a| ³ |1- a - ((-1)-a)| = 2

Hence, either |1-a| ³ 1 or |(-1)-a|³ 1.

Take any positive integer N₀ . If |a-1| ³ 1, then take any even n > N₀ and we have

|a_n - a| = |1- a| ³ 1 and if |(-1)-a|³ 1, then take any odd n > N₀ and we have |a_n - a| =|(-1)-a|³ 1.

Thus (a_n ) cannot converge to any a and so is divergent.

3. a_n = 1/n. Then a_n ® 0.

For any e > 0, there exists a positive integer N₀ such that (by the archimedean property of R). Thus for n ³ N ₀ , and this means and so by definition a_n ® 0.

We have already come across the notion of continuity and limit of a function, we shall use this notion to derive the properties of the sequence.

Let P^-1 denotes the set {1/n ; n Î P}. That is P^-1 = {1, ½, 1/3, ... }.

Then let K = P^-1 È {0} = {0, 1, ½, 1/3, ....}.

Here is an easy result:

Proposition 4. Let (a_n ) be a sequence in R. Define a function

f : K = P^-1 È {0}® R

by f (1/n) = a_n for n > 0 and f (0) = a.

Then a_n ® a if and only if f is continuous at 0.

We shall omit the proof. It is sufficient to say that this is just a restatement of the convergence of the sequence to a limit form for function.

Example 5.

1. as n ® ¥ for all k > 0.

Consider then f : K = P^-1 È {0}® R

f (1/n) = a_n = and f (0) = 0.

Thus the function is given by f (x) = x^k for x ³ 0..

This function is continuous at x = 0 by showing that its limit at 0 is 0.

since as

2. Let .

Then f : K = P^-1 È {0}® R is given by f (1/n) =

Thus f (x) = Thus since this function is continuous at 0 and f (0) . Therefore, .

Below we list the properties for sequences, some of which are easy consequences of continuity via Proposition 4.

Properties 8.

1. If a_n ® a and b_n ® b, then a_n + b_n ® a + b.

2. If a_n ® a, then la_n ® la for any real number l.

3. If a_n ® a and b_n ® b, then a_n b_n ® ab

4. If a_n ® a and a ¹ 0, then

Thus,

5. If a_n ® a and b_n ® b with b ¹ 0 , then

6. Comparison Test

If there exists a sequence ( b_n ) such that

(1) b_n ® 0,

(2) |a_n - a| £ |b_n|

Then a_n ® a.

Proof. Given e > 0, by (1), there exists an integer N such that n ³ N Þ |b_n| < e. Therefore, for all n ³ N, |a_n - a| £ |b_n| < e . This means a_n ® a .

Example.

If |a| < 1, then the sequence ( aⁿ ) converges to 0

Since |a| < 1, 1/|a| > 1. Then we can write 1/|a| = 1+ b and b > 0.

Hence

The last inequality is because for positive integer n.

Since , Thus by the Comparison Test aⁿ ® 0.

7. If (a_n ) converges, then (a_n ) is bounded.

Proof.

(a_n ) converges means there exists an a such that a_n ® a. Thus by the definition of convergence, taking e =1, there exists an integer N such that

n ³ N Þ |a_n - a|< e =1

Hence n ³ N Þ |a_n |< |a|+1

Let M = max{|a₁|, |a₂|, ¼ , |a_N-1|, |a|+1}. Then obviously |a_n |£ M for all positive integer n. This means (a_n ) is bounded.

8. If a_n ® a and b_n ® b and there exists an integer N such that a_n £ b_n for all n ³ N, then a £ b.

9. Squeeze Theorem.

If a_n ® a and b_n ® a and there exists an integer N such that for all n ³ N, a_n £ c_n £ b_n , then c_n ® a

Definition 10. A real sequence (a_n ) is increasing if n > m Þ a_n ³ a_m .

It is decreasing if n > m Þ a_n £ a_m .

It is strictly increasing if n > m Þ a_n > a_m .

It is strictly decreasing if n > m Þ a_n < a_m .

It is a monotone sequence if it is either increasing or decreasing.

Proposition 11. Suppose (a_n ) is a real bounded monotone sequence. Then (a_n ) is convergent.

Proof is omitted. Actually the statement is equivalent to the completeness of R.

Example. is a bounded increasing sequence and so is convergent

Another equivalent statement involves the notion of a Cauchy sequence. This expresses that when a sequence is somehow "bunched" together then it must be convergent.

Definition 12. (a_n ) is a Cauchy sequence if and only if given any e > 0, there exists an integer N such that for all n, m ³ N, |a_n - a_m | < e .

An easy consequence of the definition is

Any Cauchy sequence is bounded.

Theorem 13. Cauchy Principle of Convergence.

A sequence (a_n ) is convergent if and only if it is Cauchy.

This is the most important theorem. The property that every Cauchy sequence is convergent is equivalent to (order) completeness of R. This gives a charaterization of completeness for R and also for Rⁿ .

Definition 14. The notion of (a_n ) tending to + ¥ means

regarding the limit as a function on P.

Similarly, limit of (a_n ) tending to - ¥ means .

The rules for functions translate to the following:

Useful results for computing limits.

Suppose (a_n ), (b_n ) are two sequences.

1. If a_n ® +¥ or a_n ® - ¥ , then

2. If a_n ® +¥ and b_n ® a, a finite, then a_n +b_n® +¥

3. If a_n ® -¥ and b_n ® a, a finite, then a_n +b_n® -¥

4. If a_n ® +¥ and b_n ® a > 0 a finite, then a_n b_n® +¥

5. If a_n ® +¥ and b_n ® a < 0 a finite, then a_n b_n® -¥

These rules are particular useful when a_n is a rational function of n.

Example

1. ( n + 1/n ) tends to + ¥

2. 5 - n + 1/2ⁿ tends to - ¥

3.

4.

Part II Series.

Definition 1. Suppose (a_n ) is a sequence.

We can form the series

a₁ + a₂ + a₃ + ¼

More specifically, an (infinite) series consists of

(1) a sequence (a_n )

(2) the sequence (s_n ) of partial sums, where

a_n is called the n-th term of the series and s_n the n-th partial sum of the series.

If (s_n ) converges to a real number S, then we say the series converges to S and we write

We usually write å a_n or a₁ + a₂ + a₃ + ¼ for the series.

Example 2. The series c + c + c + ¼ converges if and only if c = 0.

Example 3. The series

Here . Thus the n-th partial sum,

s_n = a₁ + a₂ + a₃ + ¼ + a_n

Therefore,

Example 4. Geometric Series.

converges to if |a| < 1

We begin by letting c_n = aⁿ and s_n = c_o + c₁ + ¼ + c_n .

Then if a ¹ 1

if |a| < 1

if |a| > 1, then s_n will be unbounded and so is divergent.

if a = 1, then s_n will be unbounded and so is divergent.

If a = -1, then and so ( s_n ) is divergent.

Properties for sequences can now be translated to properties for series .

Properties 7.

(1) If å a_n converges then its sum is unique.

(2) If å a_n = a and å b_n = b, then å (a_n + b_n ) = a + b.

(3) If å a_n = a, then å la_n = la.

Definition 8. å a_n is a Cauchy series if the partial sum ( s_n ) is a Cauchy sequence.

I.e., if given e > 0, there exists an integer N such that

m>n ³ N Þ |s_n - s_m | < e

This is equivalent to saying that there exists an integer N such that for all n ³ N and for all positive integer p,

Then we have the principle convergence for series

Theorem 9. å a_n is convergent if and only if å a_n is Cauchy.

The next theorem is a quick way of telling certain series is divergent.

Proposition 10. If å a_n converges, then a_n ® 0.

Proof. If å a_n converges, then å a_n is Cauchy. Then definition 8, with p =1, shows that a_n ® 0.

Exanple

å aⁿ is divergent if |a| ³1 since (aⁿ ) does not converge to 0.

Converse of Proposition 10 is false.

Counter Example

is divergent as the observation about its n-th partial sums will reveal

s₁ = 1, s₂ = 1+ 1/2 , s₄ = 1 + 1/2 + ( 1/3+1/4) > 1 + 1/2 + 1/2

s₈ = 1 + 1/2 + ( 1/3+1/4) + (1/2+1/6+1/7+1/8) > 1 + 1/2 + 1/2 + 1/2

....................

and so (s_n ) is unbounded and so is divergent.

Now we have some nice result, a consequence of the monotone convergence theorem.

Proposition 11. å a_n is a series of real non-negative terms. Then å a_n is convergent if and only if (s_n ) is bounded.

Proposition 12 (Comparison Test)

Let å a_n and å b_n be two series of real non-negative terms such that

a_n £ b_n.

Then (1) å a_n converges if å b_n is convergent

(2) å b_n diverges if å a_n is divergent.

Example 13. is convergent.

Since and is convergent, so

is convergent . Therefore, is convergent.

Proposition 14. Suppose å |a_n| is convergent. Then å a_n converges.

Proof is just simply observing that if å |a_n| is Cauchy, then so is å a_n .

This follows from the following inequality and Definition 8 :

The converse is not true.

Definition 15. We say the series å a_n converges absolutely if å |a_n| is convergent

Example 16. is convergent but not absolutely. (converges by Alternating series test.)

Proposition 17. Suppose (a_n ) is a bounded sequence. Then converges.

Example 18. is absolutely convergent for any x.

Proposition 20 (Alternating Series Test, Leibnitz's Test)

If (a_n ) is a monotone decreasing, non-negative sequence and a_n ® 0, then å (-1)ⁿ⁺¹a_n is convergent.

The proof consists in showing that å (-1)ⁿ⁺¹a_n is Cauchy and is omitted. There is also a proof making use of the fact that s_2n = s_2n-1 - a_2n , both (s_2n ) and (s_2n-1) are bounded and monotone and so are convergent and that a_2n ® 0.

Theorem 21 (Ratio Test, D'Alembert's Test)

Suppose the series å a_n is such that exists and is equal to a.

Then (i) a < 1 implies that å a_n is absolutely convergent (hence convergent)

(ii) a > 1 implies that å a_n is divergent

(ii) a = 1. Then å a_n may converge or diverge. No inference can be made. The convergence may be investigated by other methods.

Proof is omitted.

Example å 1/n is divergent and å 1/n² is convergent. Ratio test for both gives a as 1.

Example 22.

1. is convergent as

2. for x > 0. Let a_n = n²xⁿ . Then

Thus is convergent for 0< x < 1. It is divergent for x > 1. For x =1 it is divergent.

3. (s > 0)

The series converges if s > 1; diverges if s £ 1.

Example 23. is convergent .

This is because for n > 0 , eⁿ > 1+ n + n² /2 + n³ / 6 > n³ / 6 and so

.

Therefore by the Comparison Test is convergent because is convergent by (3) above.