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)
Let us consider a sequence of natural numbers
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
which x_{n} terms are numbered with increasing natural numbers n. The two typical examples are:
where the difference between successive terms is constant - r.
where the ratio between successive terms is constant - q.
In many other cases, a form of x_{n} remains unknown.
A constant number A is the limit of a sequence {x_{n}} if for all positive small enough ε exists such a N number that all values of this sequence for which n > N are
or we can write
Note. A value of N depends how the ε number is chosen. Thus we should write rather N_{ε} than N.
Let a sequence {x_{n}} have its limit A. We choose p < A (or q > A) and ε > 0 such that
Let ε be less than A - p (or q - A). From the limit definition there is such a number N that for
is fulfilled
and the same
Thus we can write that: If a sequence {x_{n}} 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:
If a sequence {x_{n}} tends to the limit A ≠ 0 then sequence terms distant enough are
where r is a positive number. Actually, when a > 0 (< 0) then
and put r = p (or r = |q|).
If a sequence {x_{n}} has the limit A then the sequence is limited in the meaning that all its terms are
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
or
Can it be false when n ≤ N? If we take M as the greatest number of
then for all terms x_{n} we have
what should be shown.
A sequence {x_{n}} cannot have two different limits.
Let a sequence tend to A and B when A < B. Take a number r such that
Because x_{n} → A and A < r there is an index N' that for n > N'
On the other hand, because x_{n} → B and B > r there is also an index N'' that for n > N''
If we take an index n that is greater than N' and N'' then there is such a sequence term x_{n} that is lower and greater than r at the same time. It means that a sequence cannot tend to two different limits.
One can say that a sequence {x_{n}} 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 x_{n} are greater than E (or lower than -E). In other words, there is such an index N_{E} that for n > N_{E} is
One of consequences of infinity limit (positive or negative) is that terms of the sequence starting from some point are
If absolute values of terms of a sequence {x_{n}} tend to infinity then their reverse values α_{n} = 1/x_{n} tend to zero.
Let us take a number ε > 0. Because |x_{n}| → + ∞, there is such an index N that
Then for the same n values we have
what should be shown.
If terms of a sequence {x_{n}} (having values greater than zero) tend to zero then their reverse values α_{n} = 1/x_{n} tend to + ∞.
If two sequences x_{n} and y_{n} are equal i. e. if x_{n} = y_{n} and one of them has a finite limit a then the second one has also a finite limit b
and
If two sequences x_{n} and y_{n} fulfill
and each of them has a finite limit
then
Let us assume that a < b and take a number r such that
Now we can find an index N' that for n > N'
and on the other hand we can find an index N'' that for n > N''
Thus if we take N greater that both N' and N'' then for n > N we have at the same time
that gives
which is opposite to the assumption.
Note. The strong relation for sequences i. e. x_{n} > y_{n} (in many cases) leads to weak relation for the limits i. e. lim x_{n} ≥ lim y_{n}.
If sequences {x_{n}}, {y_{n}}, {z_{n}} fulfilled
and sequences {x_{n}} and {z_{n}} have the common limit a
then the sequence {y_{n}} have the same limit
Let us choose a number ε > 0. For this number one can find such an index N' that for n > N' is
Next, one can find such an index N'' that for n > N'' is
Finally, let N be greater than N' and N''. Then for n > N is
Thus for n > N
which means that
One can conclude from this that if for all n the following relation is fulfilled
and z_{n} → a then y_{n} → a.
The sum of finite number of sequences that tend to zero also tends to zero.
Proof.Let us take two sequences {x_{n}} and {y_{n}} tending to zero. Then we assume a number ε > 0. Because the sequence x_{n} has the limit zero we have that for n > N' is
We can write the same for the second sequence y_{n}
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
It gives that the sequence {x_{n} + y_{n}} tends to zero.
Lemma 2.Let us take two sequences {x_{n}} and {y_{n}}. The first sequence is limited i. e.
and the second one tends to zero. Then the sequence being their product tends to zero.
Proof.Let for all n be
Let us take a number ε > 0. We can find such an index N that for n > N is
Then we have for the same values of n fulfilled
It gives that the sequence {x_{n} y_{n}} tends to zero.
If sequences {x_{n}} and {y_{n}} have finite limits
their sum (subtraction) also has finite limit
From the theorem we have
where {α_{n}} and {β_{n}} are sequences tending to zero. In turn, we can write that
Then the sequence
tends to zero on the basis of the Lemma 1. It gives that the sequence
has the limit a ± b.
If sequences {x_{n}} and {y_{n}} have finite limits
their product has also finite limit and
From the theorem we have
The expression in brackets tends to zero. It gives that the sequence {x_{n} y_{n}} has the limit ab.
If sequences {x_{n}} and {_{n}} have finite limits
when b ≠ 0 their ratio has also finite limit
From the theorem we have
The expression in brackets tends to zero. And
because b ≠ 0 from some point is
where r is a constant number. It means that the whole expression on rhs tends to zero. Finally we have
Symbols having not – well defined meaning are the following expressions:
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.A sequence {x_{n}} increases if
i. e. if for n' > n we have x_{n'} > x_{n}.
A sequence {x_{n}} never decreases if
i. e. if for n' > n we have x_{n'} ≥ x_{n}.
A sequence {x_{n}} decreases if
i. e. if for n' > n we have x_{n'} < x_{n}.
A sequence {x_{n}} never increases if
i. e. if for n' > n we have x_{n'} ≤ x_{n}.
Let us consider increasing sequence that {x_{n}}. It has finite limit if that sequence is limited from its up i. e.
in opposite case tends to +∞.
Let us consider decreasing sequence that {x_{n}}. It has finite limit if that sequence is limited from its down i. e.
in opposite case tends to -∞.
Let us assume that sequence {x_{n}} is limited from its up. It can be written as
Then we have
and for any ε number there is such an index N that
Because the sequence {x_{n}} is monotonic increasing for n > N there is
it gives that
and
It means that lim x_{n} = a.
On the other hand, if an increasing sequence {x_{n}} is not limited from its up then for any big enough number E > 0 there is at least one index N for which
Because the sequence is increasing for n > N we have
it gives that lim x_{n} = +∞.
A sequence {x_{n}} has finite limit if and only if for every ε > 0 there is such an index N that
is fulfilled if only n > N and n' > N.
Proof. Necessary condition.Let {x_{n}} sequence has defined finite limit a. From limit definition we have that
for n > N. Let us take one more index n' > N. Then
and it gives that |x_{n} - x_{n'}| must be less than ε
Let us divide set of real numbers into two subsets. Numbers belonging to the first subset fulfill the condition
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
It is clear that each number: x_{n'} - ε (for n' > N) belongs to the first subset and each number: x_{n'} + ε (for n' > N) belongs to the second subset.
From the Dedekind's theorem we have that there is such a real number Number_{div} which fulfills the relation
Thus
for n' > N. From the limit definition we get that
Putting a = Number_{div} we end the proof of theorem.
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 {x_{n}} 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 [a_{1}, b_{1}]) must contain infinity terms of the sequence x_{n}. By analogy, the range [a_{1}, b_{1}] is divided into two equal parts and at least one of parts must contain infinity numbers of sequence x_{n}. This procedure we do infinity times. The length of k-th sub-range equals
and tends to zero with k. It means (from the previous theorem) that both b_{k} and a_{k} tend to the same limit c. Now, let us construct sub-sequence {x_{n}_{k}} in the following way:
It is possible because each compartment [a_{k}, b_{k}] has infinity numbers, thus, should have terms of this sequence having great enough indices. Then we have that
which gives that
If you want to exercise some analytic examples of finding sequence limit go to the page sequence limit - analytic examples.
If you want to learn how to find the limit of a function go to the page function - limit.