Courses in Math - Sequence - Analytic Examples

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

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

    |αn| < 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 fulfil

    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 {αn} and {βn} are sequences tending to zero. In turn, we can write that

    xn ± yn = (a ± b) + (αn ± βn).

    Then the sequence

    {αn ± βn}

    tends to zero on the basis of the Meat 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 fulfil 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 fulfils 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:

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.