##### Document Text Contents

Page 1

Mathematics (M101)

Sequence and Series

B.Tech 1st Year 1st Semester

Sequence

Definition: An Infinite sequence is a function f(n) whose domain is an infinite subset of

whole numbers.

We usually write a sequence {an} . where f(n) = an

The following are the example of sequences

{(.5)n} {n/(n+1)} {en/n!} and {(-1)n}

You'll notice that sequences are discontinuous everywhere but never the less play an

important role in Mathematics as we will see when we consider Infinite Series.

1. an = {(.5)n}

2. an = n/ (n+1)

3. an = {en/n!}

4. an= {(-1) n }

As you could tell from the examples the 1st and 3d converge to 0, the second converges

to 1 and the 4th diverges.

But do we mean by convergence exactly?

Definition:For any ε>0 there exists a number M such that L- ε < an < L + ε.

whenever n> M Or as is usually stated given any ε>0 there exists a number M such

that | an - L | < ε whenever n > M.Then the sequence an is said to converge to L.

How do we compute limits?

There are 2 very important theorems

1. Given {an} if there is a continuous differentiable function f(x) such that f(n) = an

then lim an = lim f(x) therefore we can use our results on differentiable functions and

L'Hopital's Rule.

Examples 1 and 2 fit into this category. But what about the 3d? the Gamma Function

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notwithstanding we need the following

2. If {an} is increasing and bounded above or decreasing and bounded below it

converges.

The first 3 examples make this fairly obvious.

Theorem: Suppose {

a

n

} is a sequence which is increasing and bounded above, then it

converges.

Proof: Let L be the Least Upper Bound of {

a

n

} i.e. it is an upper bound and there are

none smaller.

Let ε be given.

Let M be the first number such that

a

m

> L - ε . Such a number exists otherwise L -ε

would be an Upper Bound.

For all n>M

a

n

>

a

m

since

a

n

is an increasing sequence.

Further for all n

a

n

< L + ε because if L is an upper bound L+ ε is an upper bound.

Therefore we have shown given any ε > 0 There is a number M such that

L - ε <

a

n

< L+ ε whenever n > M i.e {

a

n

} converges --In fact it converges to its

Least Upper Bound.

Alternating Sequences:

Another important theorem

Suppose bn = (-1)n an where an >0. Then if lim an = 0 then lim bn = 0

If lim an ≠ 0 then { bn} diverges. Even if lim an = L The alternating sequence (-1)n an

diverges as the subsequence of even terms converges to L and the subsequence of odd

terms converges to - L therefore the sequence diverges.

The next 2 examples show the divergent case and the 3d shows convergence of an

alternating sequence.

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diverges.

D’Alembert’s Ratio Test: Let ∑an be a series, and assume that

1lim n

x

n

u

l

u

+

→∞

= . Then if l<

1, the series is convergent, if l > 1, the series is divergent, while if l = 1, the test gives no

information.

Example: Let an = . Then ∑an is convergent.

Solution. We look the ratio of adjacent terms in the series (of positive terms).

= =

= = as n .

Since the ratio of adjacent terms in the series tends to a limit which is < 1, the

series converges by the ratio test.

Example: Determine if the following series is convergent or divergent.

Solution

Here’s the limit.

Again, the ratio test tells us nothing here. We can however, quickly use the divergence

test on this. In fact that probably should have been our first choice on this one anyway.

By the Divergence Test this series is divergent.

Mathematics (M101)

Sequence and Series

B.Tech 1st Year 1st Semester

Sequence

Definition: An Infinite sequence is a function f(n) whose domain is an infinite subset of

whole numbers.

We usually write a sequence {an} . where f(n) = an

The following are the example of sequences

{(.5)n} {n/(n+1)} {en/n!} and {(-1)n}

You'll notice that sequences are discontinuous everywhere but never the less play an

important role in Mathematics as we will see when we consider Infinite Series.

1. an = {(.5)n}

2. an = n/ (n+1)

3. an = {en/n!}

4. an= {(-1) n }

As you could tell from the examples the 1st and 3d converge to 0, the second converges

to 1 and the 4th diverges.

But do we mean by convergence exactly?

Definition:For any ε>0 there exists a number M such that L- ε < an < L + ε.

whenever n> M Or as is usually stated given any ε>0 there exists a number M such

that | an - L | < ε whenever n > M.Then the sequence an is said to converge to L.

How do we compute limits?

There are 2 very important theorems

1. Given {an} if there is a continuous differentiable function f(x) such that f(n) = an

then lim an = lim f(x) therefore we can use our results on differentiable functions and

L'Hopital's Rule.

Examples 1 and 2 fit into this category. But what about the 3d? the Gamma Function

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http://calculus7.com/sitebuildercontent/sitebuilderfiles/enbyn.avi

http://calculus7.com/sitebuildercontent/sitebuilderfiles/nbynplus1.avi

http://calculus7.com/sitebuildercontent/sitebuilderfiles/halfton.avi

Page 2

notwithstanding we need the following

2. If {an} is increasing and bounded above or decreasing and bounded below it

converges.

The first 3 examples make this fairly obvious.

Theorem: Suppose {

a

n

} is a sequence which is increasing and bounded above, then it

converges.

Proof: Let L be the Least Upper Bound of {

a

n

} i.e. it is an upper bound and there are

none smaller.

Let ε be given.

Let M be the first number such that

a

m

> L - ε . Such a number exists otherwise L -ε

would be an Upper Bound.

For all n>M

a

n

>

a

m

since

a

n

is an increasing sequence.

Further for all n

a

n

< L + ε because if L is an upper bound L+ ε is an upper bound.

Therefore we have shown given any ε > 0 There is a number M such that

L - ε <

a

n

< L+ ε whenever n > M i.e {

a

n

} converges --In fact it converges to its

Least Upper Bound.

Alternating Sequences:

Another important theorem

Suppose bn = (-1)n an where an >0. Then if lim an = 0 then lim bn = 0

If lim an ≠ 0 then { bn} diverges. Even if lim an = L The alternating sequence (-1)n an

diverges as the subsequence of even terms converges to L and the subsequence of odd

terms converges to - L therefore the sequence diverges.

The next 2 examples show the divergent case and the 3d shows convergence of an

alternating sequence.

Page 13

diverges.

D’Alembert’s Ratio Test: Let ∑an be a series, and assume that

1lim n

x

n

u

l

u

+

→∞

= . Then if l<

1, the series is convergent, if l > 1, the series is divergent, while if l = 1, the test gives no

information.

Example: Let an = . Then ∑an is convergent.

Solution. We look the ratio of adjacent terms in the series (of positive terms).

= =

= = as n .

Since the ratio of adjacent terms in the series tends to a limit which is < 1, the

series converges by the ratio test.

Example: Determine if the following series is convergent or divergent.

Solution

Here’s the limit.

Again, the ratio test tells us nothing here. We can however, quickly use the divergence

test on this. In fact that probably should have been our first choice on this one anyway.

By the Divergence Test this series is divergent.