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TitleSequence and Series
TagsSequence Summation Series (Mathematics) Monotonic Function Limit (Mathematics)
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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.

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.

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