# Courses in Math - Sequence - Analytic Examples

This page helps you learn analytic methods of finding limits of sequences. One can become familiar with elementary theorems of theory of sequences.

Literature.

G. M. Fichtenholz – Integral and differential calculus, vol. 1, PWN Warsaw 1999 (originally in Russian)

## Sequence. Definition.

Let us consider a sequence of natural numbers

1, 2, 3, … n, …,

in which numbers preserve growing order (the sequence is monotonically increasing). Replacing each natural number in this sequence by a real number leads to a real numbers sequence

x1, x2, x3, … xn, …,

which xn terms are numbered with increasing natural numbers n. The two typical examples are:

1. ## Arithmetic sequence

a, a + r, a + 2r, … a + r(n - 1), …,

where the difference between successive terms is constant - r.

2. ## Geometric sequence

a, aq, aq2, … aqn-1, …,

where the ratio between successive terms is constant - q.

In many other cases, a form of xn remains unknown.

## Limit Of A Sequence.

A constant number A is the limit of a sequence {xn} if for all positive small enough ε exists such a N number that all values of this sequence for which n > N are

|xn - A| < ε or A - ε < xn < A + ε

or we can write

lim xn = A or xn → A

Note. A value of N depends how the ε number is chosen. Thus we should write rather Nε than N.

## Theorem Of A Sequence Having The Limit.

Let a sequence {xn} have its limit A. We choose p < A (or q > A) and ε > 0 such that

A - ε > p or (A + ε < q).

Let ε be less than A - p (or q - A). From the limit definition there is such a number N that for

n > N

is fulfilled

xn > A - ε (xn < A + ε),

and the same

xn > p (xn < q).

Thus we can write that: If a sequence {xn} tends to A and A > p (or A < q) then all terms of this sequence with indices greater than an index n' are greater than p (or lower than q).

This theorem has a lot of implications:

1. If a sequence {xn} tends to the limit A > 0 (or < 0) then a term xn > 0 (or < 0) starting from a some point. (put p = 0 or q = 0, respectively)
2. If a sequence {xn} tends to the limit A ≠ 0 then sequence terms distant enough are

|xn| > r > 0 (for n > N).

where r is a positive number. Actually, when a > 0 (< 0) then

0 < p < A (A < q < 0)

and put r = p (or r = |q|).

3. If a sequence {xn} has the limit A then the sequence is limited in the meaning that all its terms are

|xn| ≤ M (where M = const; n = 1, 2, 3, …).

Let a number M' > |A| such that -M' < A < M' and let us put p = -M' and q = M'. There is an index N that when n > N is

-M' < xn < M'

or

|xn| < M' when n = N + 1, N + 2, ….

Can it be false when n ≤ N? If we take M as the greatest number of

|x1|, |x2|, …, |xN|, M'

then for all terms xn we have

|xn| < M'

what should be shown.

4. A sequence {xn} cannot have two different limits.

Let a sequence tend to A and B when A < B. Take a number r such that

A < r < B.

Because xn → A and A < r there is an index N' that for n > N'

xn < r.

On the other hand, because xn → B and B > r there is also an index N'' that for n > N''

xn > r.

If we take an index n that is greater than N' and N'' then there is such a sequence term xn that is lower and greater than r at the same time. It means that a sequence cannot tend to two different limits.

### A Sequence Having Infinity Limit.

One can say that a sequence {xn} has the limit +∞ (-∞) or equivalently that terms of a sequence diverge to +∞ (-∞) if for a big enough number E > 0 starting from some point terms of sequence xn are greater than E (or lower than -E). In other words, there is such an index NE that for n > NE is

xn > E (or respectively xn < -E).

One of consequences of infinity limit (positive or negative) is that terms of the sequence starting from some point are

xn > 0 (or respectively xn < 0).

#### Theorems.

1. If absolute values of terms of a sequence {xn} tend to infinity then their reverse values αn = 1/xn tend to zero.

Let us take a number ε > 0. Because |xn| → + ∞, there is such an index N that

|xn| > 1/ε if only n > N.

Then for the same n values we have

n| < ε,

what should be shown.

2. If terms of a sequence {xn} (having values greater than zero) tend to zero then their reverse values αn = 1/xn tend to + ∞.

### Theorems useful in Finding Limit.

Note. The below – presented theorems helps to find the limit of a sequence being in some kind of relation to another sequence having limit.
1. If two sequences xn and yn are equal i. e. if xn = yn and one of them has a finite limit a then the second one has also a finite limit b

lim xn = a, lim yn = b

and

a = b.
2. If two sequences xn and yn fulfill

xn ≥ yn

and each of them has a finite limit

lim xn = a, lim yn = b

then

a ≥ b.

Let us assume that a < b and take a number r such that

a < r < b.

Now we can find an index N' that for n > N'

xn < r

and on the other hand we can find an index N'' that for n > N''

yn > r.

Thus if we take N greater that both N' and N'' then for n > N we have at the same time

xn < r, yn > r,

that gives

xn < yn

which is opposite to the assumption.

Note. The strong relation for sequences i. e. xn > yn (in many cases) leads to weak relation for the limits i. e. lim xn ≥ lim yn.

3. If sequences {xn}, {yn}, {zn} fulfilled

xn ≤ yn ≤ zn

and sequences {xn} and {zn} have the common limit a

lim xn = lim zn = a,

then the sequence {yn} have the same limit

lim yn = a.

Let us choose a number ε > 0. For this number one can find such an index N' that for n > N' is

a - ε < xn < a + ε.

Next, one can find such an index N'' that for n > N'' is

a - ε < zn < a + ε.

Finally, let N be greater than N' and N''. Then for n > N is

a - ε < xn ≤ yn ≤ zn < a + ε.

Thus for n > N

a - ε < yn < a + ε

which means that

lim yn = a.

One can conclude from this that if for all n the following relation is fulfilled

a ≤ yn ≤ zn

and zn → a then yn → a.

4. #### Lemma of sequences tending to zero.

Lemma 1.

The sum of finite number of sequences that tend to zero also tends to zero.

Proof.

Let us take two sequences {xn} and {yn} tending to zero. Then we assume a number ε > 0. Because the sequence xn has the limit zero we have that for n > N' is

|xn| < 1/2 ε.

We can write the same for the second sequence yn

|yn| < 1/2 ε

that is true for n > N''. Then we take a number N that is greater than both N' and N'' and for indices n > N we have

|xn + yn| ≤ |xn| + |yn| < 1/2 ε + 1/2 ε = ε

It gives that the sequence {xn + yn} tends to zero.

Lemma 2.

Let us take two sequences {xn} and {yn}. The first sequence is limited i. e.

|xn| < 1/2 ε.

and the second one tends to zero. Then the sequence being their product tends to zero.

Proof.

Let for all n be

|xn| < M.

Let us take a number ε > 0. We can find such an index N that for n > N is

|yn| < ε/M.

Then we have for the same values of n fulfilled

|xn yn| = |xn| |yn| < M ε/M = ε.

It gives that the sequence {xn yn} tends to zero.

5. #### Arithmetic operations on sequences. Limit.

##### Sum (subtraction) of sequences. Limit.

If sequences {xn} and {yn} have finite limits

lim xn = a, lim yn = b

their sum (subtraction) also has finite limit

lim (xn ± yn) = a ± b.
Proof.

From the theorem we have

xn = a + αn; yn = b + βn

where n} and n} are sequences tending to zero. In turn, we can write that

xn ± yn = (a ± b) + (αn ± βn).

Then the sequence

n ± βn}

tends to zero on the basis of the Lemma 1. It gives that the sequence

{xn ± yn}

has the limit a ± b.

##### Product of sequences. Limit.

If sequences {xn} and {yn} have finite limits

lim xn = a; lim yn = b.

their product has also finite limit and

lim (xn yn) = ab.
Proof.

From the theorem we have

xn yn = ab + (a βn + b αn + αn βn).

The expression in brackets tends to zero. It gives that the sequence {xn yn} has the limit ab.

##### Ratio of sequences. Limit.

If sequences {xn} and {n} have finite limits

lim xn = a; lim yn = b.

when b ≠ 0 their ratio has also finite limit

lim (xn/yn) = a/b.
Proof.

From the theorem we have

xn/yn - a/b = (a αn)/(b + βn) - a/b = 1/(b yn)(b αn - a βn).

The expression in brackets tends to zero. And

|1/(b yn)| = 1/|b| 1/|yn)| < 1/|b| 1/r

because b ≠ 0 from some point is

|yn| > r > 0,

where r is a constant number. It means that the whole expression on rhs tends to zero. Finally we have

xn/yn = a/b.

### Sequence. Undefined expressions.

Symbols having not – well defined meaning are the following expressions:

xn/yn → ∞/∞; 0/0
xn - yn → ∞ - ∞
xn yn → 0 ∞

The Stolz theorem helps to find the limit of such expressions.

Note. The below – presented theorems helps to find the limit of a sequence having some properties. There is no correspondence to another sequence.

### Monotonic Sequence. Definitions.

#### Increasing sequence.

A sequence {xn} increases if

x1 < x2 < … < xn < xn + 1 < …

i. e. if for n' > n we have xn' > xn.

#### Never decreasing sequence.

A sequence {xn} never decreases if

x1 ≤ x2 ≤ … ≤ xn ≤ xn + 1 ≤ …

i. e. if for n' > n we have xn' ≥ xn.

#### Decreasing sequence.

A sequence {xn} decreases if

x1 > x2 > … > xn > xn + 1 > …

i. e. if for n' > n we have xn' < xn.

#### Never increasing sequence.

A sequence {xn} never increases if

x1 ≥ x2 ≥ … ≥ xn ≥ xn + 1 ≥ …

i. e. if for n' > n we have xn' ≤ xn.

#### Monotonic sequence. Limit - theorem.

##### Increasing sequence. Limit.

Let us consider increasing sequence that {xn}. It has finite limit if that sequence is limited from its up i. e.

xn ≤ M (M = const; n = 1, 2, 3, …)

in opposite case tends to +∞.

##### Decreasing sequence. Limit.

Let us consider decreasing sequence that {xn}. It has finite limit if that sequence is limited from its down i. e.

xn ≥ m (m = const; n = 1, 2, 3, …)

in opposite case tends to -∞.

##### Proof (increasing sequence).

Let us assume that sequence {xn} is limited from its up. It can be written as

a = sup {xn}.

Then we have

xn ≤ a.

and for any ε number there is such an index N that

xN > a - ε.

Because the sequence {xn} is monotonic increasing for n > N there is

xn > xN

it gives that

xn > a - ε.

and

0 ≤ a - xn < ε → |xn - a| < ε.

It means that lim xn = a.

On the other hand, if an increasing sequence {xn} is not limited from its up then for any big enough number E > 0 there is at least one index N for which

xN > E.

Because the sequence is increasing for n > N we have

xn > E

it gives that lim xn = +∞.

### Sequence. Limit Point.

#### Cauchy - Bolzano criterion. Theorem.

A sequence {xn} has finite limit if and only if for every ε > 0 there is such an index N that

|xn - xn'| < ε

is fulfilled if only n > N and n' > N.

Proof. Necessary condition.

Let {xn} sequence has defined finite limit a. From limit definition we have that

|xn - a| < ε/2

for n > N. Let us take one more index n' > N. Then

|xn - a| < ε/2
|xn' - a| = |a - xn'| < ε/2

and it gives that |xn - xn'| must be less than ε

|xn - xn'| = |(xn - a + a - xn'| ≤ |xn - a| + |a - xn'| < ε/2 + ε/2 = ε
Sufficient condition.

Let us divide set of real numbers into two subsets. Numbers belonging to the first subset fulfill the condition

xn > Numbersubset1

from some point. The rest of real numbers belongs to the second subset.

Let us take ε > 0 and corresponding to it index N. If n > N and n' > N then we have

xn' - ε < xn < xn' + ε.

It is clear that each number: xn' - ε (for n' > N) belongs to the first subset and each number: xn' + ε (for n' > N) belongs to the second subset.

From the Dedekind's theorem we have that there is such a real number Numberdiv which fulfills the relation

Numbersubset1 ≤ Numberdiv ≤ Numbersubset2.

Thus

xn' - ε ≤ Numberdiv ≤ xn' + ε
|Numberdiv - xn'| = |xn' - Numberdiv| ≤ ε

for n' > N. From the limit definition we get that

Numberdiv = lim xn.

Putting a = Numberdiv we end the proof of theorem.

### Bolzano - Weierstrass Lemma.

From any limited sequence (i. e. all numbers from that sequence belong to some range [a, b]) can be taken a sub-sequence that has its limit.

Proof.

Let all numbers from {xn} sequence belong to the range [a, b]. Next, let us divide this range into two equal parts. Then at least one half of the range [a, b] (denoted as [a1, b1]) must contain infinity terms of the sequence xn. By analogy, the range [a1, b1] is divided into two equal parts and at least one of parts must contain infinity numbers of sequence xn. This procedure we do infinity times. The length of k-th sub-range equals

bk - ak = [b - a]/2k

and tends to zero with k. It means (from the previous theorem) that both bk and ak tend to the same limit c. Now, let us construct sub-sequence {xnk} in the following way:

• as the first term of the new sequence xn1 we take any number from the sequence xn being in the range [a1, b1],
• as the second term of the new sequence xn2 we take any number from the sequence xn having index greater than xn1 and lying in the range [a2, b2], etc.

It is possible because each compartment [ak, bk] has infinity numbers, thus, should have terms of this sequence having great enough indices. Then we have that

ak ≤ xnk ≤ bk
lim ak = lim bk = c

which gives that

lim xnk = c.

### Sequence Limit - Analytic Examples.

If you want to exercise some analytic examples of finding sequence limit go to the page sequence limit - analytic examples.

### Function. Limit.

If you want to learn how to find the limit of a function go to the page function - limit.