By limes superior and limes inferior, mathematicians denote the largest and the smallest accumulation point of a sequence. They are useful, if there are multiple limits and intuitively say what the "greatest limit" (limes superior) and the "smallest limit" (limes inferior) of that sequence are.
We already learned about limits of a sequence. A limit is that unique number, to which a sequence tends. In every neighbourhood of the sequence, there are almost all elements, meaning only a finite number is allowed to be on the outside:
Sometimes, it seems like a sequence tends towards multiple numbers (like "multiple limits"). We also discussed that case: these numbers are then called "accumulation points" instead of "limits", since a limit must always be unique.
The set of accumulation points may be bounded or unbounded. In case it is bounded, there is a best upper and a best lower limit for the accumulation points, which we will call limes superior and limes inferior. Both are real numbers. Mathematically, for a sequence we will denote the limes superior as and the limes inferior as .
The closed interval then includes all accumulation points. We can even show that in any neighbourhood of this interval (i.e. a slightly bigger interval), there are almost all elements inside this neighbourhood. The following figure illustrates this situation for some -neighbourhood around the original interval:
Now, let us turn to a mathematical description of "greatest and smallest accumulation point". We can directly define:
Definition (limes superior)
For a sequence bounded from above, the limes superior denotes the greatest accumulation point and is denoted by . If the sequence is unbounded from above or there are no accumulation points, we write .
Definition (limes inferior)
For a sequence bounded from below, the limes inferior denotes the smallest accumulation point and is denoted by . If the sequence is unbounded from below or there are no accumulation points, we write .
But: Does this definition even make sense? The accumulation points form a set. Those sets need not to have a maximum (greatest value) or minimum (smallest value), but might instead just have a supremum or infimum. Mathematicians wondered, when this is the case and soon found a surprising answer: The "awkward case" that there is no greatest/smallest limit does never occur! This statement ca actually be proven
Theorem (greatest and smallest accumulation point exist)
If the accumulation points of a sequence are bounded from above, then it has a greatest accumulation point.
If the accumulation points of a sequence are bounded from below, then it has a smallest accumulation point.
Proof (greatest and smallest accumulation point exist)
Suppose first that the accumulation points of are bounded from above. Then, there must be a supremum (smallest upper bound) for the accumulation points. If is an accumulation point, it is also the greatest one and we get no contradiction to the above statement.
So let us assume that is not an accumulation itself. In that case, each interval for must contain an accumulation point . Otherwise, would be an upper bound for the accumulation points which is smaller than (which cannot be). But since each subsequence element with . By the triangle inequality, . This holds true for any , so is an accumulation point of itself and we get a contradiction to not being an accumulation point.
This theorem establishes that the two definitions above actually make sense, so limes superior and limes inferior are well-defined.
If limes superior and limes inferior of a sequence exist and coincide, then the greatest and smallest accumulation point are identical, so there can only be one accumulation point. And the sequence cannot be unbounded, so it should converge to this one accumulation point. But does it actually do that? And does the converse hold true? I.e., if the sequence converges, are limes superior and inferior identical? The answer turns out to be yes:
Theorem (limes superior/inferior and convergence)
A sequence converges if and only if
How to get to the proof? (limes superior/inferior and convergence)
We need to prove equivalence
which are two directions:
"" is easy: if a sequence converges, then there is exactly one accumulation point (namely the limit ). This must be the greatest and smallest accumulation point at the same time, so .
"" needs some technical work, where we will use the definition of a limit and accumulation point via neighbourhoods. We recall:
A point is the limit of , if and only if there are almost all
A point is an accumulation point of , if and only if there are infinitely many.
Proof (limes superior/inferior and convergence)
Step 1: ""
If , then the sequence is bounded and is its only accumulation point. By definition of limsup and liminf:
Step 2: ""
Since is the greatest accumulation point of , we know that for all , there are infinitely many sequence elements in and almost all in (i.e. only finitely many on the outside).
Since is the smallest accumulation point of , for any , there are infinitely many sequence elements in and almost all of them in (i.e. only finitely many on the outside).
So if we fix , then only "finite + finite = finite"-ly many sequence elements can be outside . So almost all are on the inside which is equivalent to saying that converges to .
This theorem can be extended to improperly converging sequences:
Exercise (limes superior/inferior and improper convergence)
Prove that the statement in the above hint holds for . (The case is proven analogously)
How to get to the proof? (limes superior/inferior and improper convergence)
Again, we need to show two directions: " is by definition of limsup and liminf and " needs some technical work.
Proof (limes superior/inferior and improper convergence)
Step 1:. Let . Then, is unbounded from above, but bounded from below. So by definition, (and ). Further, has no accumulation point (otherwise it would not converge improperly to infinity), which implies by definition that .
Step 2:. Let . Then, is unbounded from above, but bounded from below. Since we know that has no accumulation points. We will show that this implies :
Let be any real sequence bounded from below, without an accumulation point in . Then, converges improperly to .
This is shown by contradiction: Assume that is bounded from below, but it does not converge improperly to , which means
This means, there are infinitely many with . By means of the Bolzano-Weierstrass theorem, one can find a subsequence of which converges to some . This means, is an accumulation point of , which contradicts not having any accumulation points. So any bounded from below without accumulation points must diverge improperly to . This also holds true for our above and by definition, , which is what we wanted to show.
Let . This sequence has two accumulation points: and . So by the original definition, and . In order to apply the alternative definition above, we have to determine and , as well as their limits. This is an easy task here: since only attains the values and both infinitely many, there is just and for all . So we have two constant sequences with limits
and both definitions coincide.
Consider . this is a null sequence, so by the original definition,
For the new definition of limsup, we determine
and both definitions of the limsup coincide.
Question: Analogously, determine by the new definition.
which also coincides with the original definition.
How to get to the proof? (alternative definition of limsup and liminf)
At first, we focus on the "limsup-case" and prove that indeed converges. Since This sequence is monotonously decreasing and bounded, so it converges by the monotony criterion. Then, we need to show that indeed converges to , i.e. the greatest accumulation point of . This is done by establishing the two inequalities and .
The "liminf-case" works analogously. We use that in monotonously increasing an d bounded, so it converges. And we show that the limit is indeed , meaning the smallest accumulation point of .
Proof (alternative definition of limsup and liminf)
Since is bounded (by assumption), so is . In addition,
So is monotonously decreasing. By the monotony criterion, converges to some limit which we call .
Let us denote the greatest accumulation point of by . Our aim is to show . This is done by proving and :
Since is an accumulation point of , we know that for each there is some with and
for any . This especially implies for all . Taking the limit, we obtain .
On the other hand, as is monotonously decreasing, there is also . This holds for all . Taking the limit , we get , which finishes the proof.
Rules for computing with limsup and liminf[Bearbeiten]
Theorem (monotony rule)
Let and be bounded real sequences with for all . Then
Proof (monotony rule)
Let be the greatest accumulation point of and the greatest accumulation point of . Those accumulation points exist by the Bolzano-Weierstrass theorem and by boundedness of the sequences. Let be given. since is the greatest accumulation point of , there must be an , such that for all : . But since , there is also . Therefore holds. Taking the limit , we obtain .
Analogously, prove that .
Theorem (Relation between limsup and liminf - mirroring along the x-axis)
Let be a bounded real sequence. Then,
Proof (Relation between limsup and liminf - mirroring along the x-axis)
Let be the greatest accumulation point of . We need to show that , i.e. is tha smallest accumulation point of . This is done in two steps:
We prove that is an accumulation point of
We prove that is the smallest accumulation point of
Step 1: Since , there is a subsequence of with . By the limit theorems, we have . So the subsequence of converges to , i.e. is an accumulation point of .
Step 2: Let be any accumulation point of . Then, there is a subsequence of with . By the limit theorems, . hence, is an accumulation point of . Since was the greatest accumulation point of , we know that , or equivalently .
As was any accumulation point of , we know that any accumulation point is greater or equal . So is the smallest one and .
Theorem (Sum rule)
Let and be real sequences. Then,
Proof (Sum rule)
Step 1: We start with the second inequality :
Let , and let be an accumulation point of . Then, there is a subsequence of with .
Let further be arbitrary. Since is the greatest accumulation point of and is the greatest accumulation point of , there must be a , such that for all there is: and . So, for we also have . By monotony of the limits, we know that , so the can be at most above our desired threshold . But now, can be chosen arbitrarily close to 0, so we get . Since this inequality holds for all accumulation points of , it also holds for the greatest one and
Step 2: Now, we prove . This can be done by usind the second inequality which we just have proven and the rule above. there is
This is exactly equivalent to .
Exercise (Example for the sum rule)
Find explicit sequences and with strict bounds
Solution (Example for the sum rule)
The trick is to let sequence elements cancel against each other. One needs to play a bit around with alternating sequences to get to a solution. One example is
The first sequence elements are
So both and have the three accumulation points and . The sum of both sequences is
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