The squeeze theorem is a powerful tool to determine the limit of a complicated sequence. It is based on comparison to simpler sequences, for which the limit is easily determinable.
Convergence proof for a root sequence (video in German)
The intuition behind this theorem is quite simple: We are given a complicated sequence
and want to know whether it converges. Often, one can leave out terms in the complicated sequence
and gets some simpler sequences
and
. If
is a lower bond and
an upper bound, then
is "caught" in the space between both functions. If both sequences converge to the same limit
, they "squeeze" together this space to this single point and
has no other option than to converge towards
, as well.
You may also visualize this theorem by a "ham & cheese sandwich" (see image on the right). The upper and lower bounds
and
act as two "slices of toast", which confine
, i.e. the "filling". If you squeeze the toast slices together, the filling in between will also squeezed to this point.
The squeeze theorem[Bearbeiten]
The theorem reads as follows:
Proof (Squeeze theorem)
Let
and
be such that
and
for some
.
We need to prove
, i.e. for each
there is an
, such that
for all
. Let
be arbitrary. By assumption, the convergences
and
hold.
Hence, there are two thresholds
and
with
for all
(lower bounding sequence) and
for all
(upper bounding sequence). Now, there is
for
. For each
we distinguish the two cases
(above the limit) and
(below the limit).
In case
there is
The other case
is treated analogously:
So if the maximum of both thresholds
is exceeded by
( i.e.
and
hold), then for
,
and for
,
So in either case
for all
, which establishes convergence.
Example
We illustrate the squeeze theorem using the root sequence
for an arbitrary but fixed constant
.
Two cases need to be distinguished:
Fall 1: 
Fall 2: 
A useful special case[Bearbeiten]
We often encounter 0 as a sequence limit (null sequence). Since the squeeze theorem can be used to prove any
to be a limit of a sequence
, it can also be used for
. Especially, for any convergent
, the sequence
must converge to 0:
Hint
This special case of the squeeze theorem is also referred to as direct comparison argument for sequences .
Squeeze theorem: examples and problems[Bearbeiten]
Squeeze theorem: example & exercise 1[Bearbeiten]
Example (Squeeze theorem 1)
Let us consider the sequence
with
The first 6 elements of the sequence

,

and

We take a look at the first sequence elements
. They seem to be alternately positive and negative and they tend to zero. We ise as lower or upper bound
and
. Then,
dso we caught
for all
.
In addition
. The squeeze theorem hence implies
.
We could also use
and
. Or any other power between 0 and 2 (not necessarily 1 or 2).
Exercise (Squeeze theorem 1)
Investigate whether the sequence
converges, using the squeeze theorem, where
with
Squeeze theorem: example & exercise 2[Bearbeiten]
Exercise (Squeeze theorem 2)
Use the squeeze theorem to show that
and
with


both converge to 0.
Squeeze theorem: example & exercise 3[Bearbeiten]
Example (Squeeze theorem 3)
Now, let us consider the root sequences
and
with
and
First sequence: For
, monotony of the
-th root implies
as well as
For those two bounding sequences, the limit theorems imply
and
The squeeze theorem hence implies
.
Second sequence: For
the root function is monotonous, as well
and
We use the limit theorems for the upper and lower bounding sequences
and
So the squeeze theorem implies
.
Solution (Squeeze theorem 3)
Part 1: For
the
-th root is monotonous, so
further, for
Hence, there is
The limit theorems for the bounding sequences yield
and
So the squeeze theorem implies
.
Part 2: For
we again have monotony of the
-th root
Further,
The limit theorems for the bounding sequences render
And by means of the squeeze theorem
.
Squeeze theorem: example & exercise 4[Bearbeiten]
Example (Squeeze theorem 4)
Let
be the sequence of partial sums
Then, for all
we bound:
and hence
For the lower bound,
For the upper bound,
So the squeeze theorem implies
.
Exercise (Squeeze theorem 4)
Investigate the limits of the two sequences of partial sums
and
defined by


using the squeeze theorem.
Solution (Squeeze theorem 4)
Part 1: For all
we have:
so we have some bounds
For the upper bound,
For the lower bound,
So the squeeze theorem implies
.
Part 2: For all
and
there is:
so we have the bounds
The lower bound is just 0 and for the upper bound,
So the squeeze theorem implies
.
Squeeze theorem: example & exercise 5[Bearbeiten]
Exercise (Squeeze theorem 5)
Does the series
converge`? If yes, what is its limit? Provide proofs for your claims.
How to get to the proof? (Squeeze theorem 5)
Perhaps, you have already encountered the sequence
which converges to Euler's number
. This is an upper bound for
, so with that knowledge, we may assert that it converges. But what is the limit? Maybe, we should start by computing some initial sequence elements.:
It looks as if
converges to
. This gets particularly clear, if we plot the sequence elements in a diagram:
What makes determining the limit hard is the exponent of
. Otherwise, we could use some limit theorems of an epsilon-delta proof and would be done quite fast. Luckily, there is a trick to get rid of the exponent: Bernoulli's inequality:
The sequence
is hence a lower bound on
and it clearly converges to
. We hence set
as a lower bound for the squeeze theorem.
How to get an upper bound converging to 1? This is quite easy, since
is always smaller than
: The expression
is always smaller than 1 and therefore, also every power of it will be. This includes
being smaller than
.
So for the squeeze theorem, we have for all
the inequality:
both the lower and the upper bounding sequence converge to the identical limit
. So by means of the squeeze theorem, there must also be
.
Proof (Squeeze theorem 5)
Bernoulli's inequality implies
And there is
So we get the bounds
Both bounding sequences converge to
, so by means of the squeeze theorem,
which was to be shown.
Hint
If you are allowed to use
and
, then you can also obtain the result directly, using the third binomial formula:
Exercise (Squeeze theorem 5)
Does the sequence
converge? If yes, what is its limit? Prove your assertions.
Examples & exercises: squeeze theorem for null sequences[Bearbeiten]
Exercise (Squeeze theorem for null sequences)
Prove that
.
Solution (Squeeze theorem for null sequences)
In order to use the squeeze theorem, we need to bound
from above by a null sequence. The binomial theorem can be used for this bounding. In case
:
This inequality implies
So the desired bound reads
And by means of the squeeze theorem
, which was to be shown.