# 2.5: Limit Superior and Limit Inferior - Mathematics

We begin this section with a proposition which follows from Theorem 2.3.1. All sequences in this section are assumed to be of real numbers.

Proposition (PageIndex{1})

Let (left{a_{n} ight}) be a bounded sequence. Define

[s_{n}=sup left{a_{k}: k geq n ight}]

and

[t_{n}=inf left{a_{k}: k geq n ight}.]

Then (left{s_{n} ight}) and (left{t_{n} ight}) are convergent.

Proof

If (n leq m), then (left{a_{k}: k geq m ight} subsetleft{a_{k}: k geq n ight}). Therefore, it follows from Theorem 2.5.3 that (s_{n} geq s_{m}) and, so, the sequence (left{s_{n} ight}) is decreasing. Since (left{a_{n} ight}) is bounded, then so is (left{s_{n} ight}). In particular, (left{s_{n} ight}) is bounded below. Similarly, (left{t_{n} ight}) is increasing and bounded above. Therefore, both sequences are convergent by Theorem 2.3.1. (square)

Definition (PageIndex{1}): Limit Superior

Let (left{a_{n} ight}) be a sequence. Then the limit superior of (left{a_{n} ight})), denoted by (limsup _{n ightarrow infty} a_{n}), is defined by

[limsup _{n ightarrow infty} a_{n}=lim _{n ightarrow infty} sup left{a_{k}: k geq n ight}.]

Note that (limsup _{n ightarrow infty} a_{n}=lim _{n ightarrow infty} s_{n}), where (s_{n}) is defined in (2.8).

Similarly, the limit inferior of (left{a_{n} ight}), denoted by (liminf _{n ightarrow infty} a_{n}), is defined by

[liminf _{n ightarrow infty} a_{n}=liminf _{n ightarrow infty}left{a_{k}: k geq n ight}.]

Note that (liminf _{n ightarrow infty} a_{n}=lim _{n ightarrow infty} t_{n}), where (t_{n}) is defined in (2.9).

Theorem (PageIndex{2})

If (left{a_{n} ight}) is not bounded above, then

[lim _{n ightarrow infty} s_{n}=infty,]

where (left{s_{n} ight}) is defined in (2.8).

Similarly, if (left{a_{n} ight}) is not bounded below, then

[lim _{n ightarrow infty} t_{n}=-infty,]

where (left{t_{n} ight}) is defined in (2.9).

Proof

Suppose (left{a_{n} ight}) is not bounded above. Then for any (k in mathbb{N}), the set (left{a_{i}: i geq k ight}) is also not bounded above. Thus, (s_{k}=sup left{a_{i}: i geq k ight}=infty) for all (k). Therefore, (lim _{k ightarrow infty} s_{k}=infty). The proof for the second case is similar. (square)

Remark (PageIndex{3})

By Theorem 2.5.2, we see that if (left{a_{n} ight}) is not bounded above, then

[limsup _{n ightarrow infty} a_{n}=infty.]

Similarly, if (left{a_{n} ight}) is not bounded below, then

[liminf _{n ightarrow infty} a_{n}=-infty.]

Theorem (PageIndex{4})

Let (left{a_{n} ight}) be a sequence and (ell in mathbb{R}). The following are equivalent:

1. (limsup _{n ightarrow infty} a_{n}=ell).
2. For any (varepsilon>0), there exists (N in mathbb{R}) such that

[a_{n}

and there exists a subsequence of (left{a_{n_{k}} ight}) of (left{a_{n} ight}) such that

[lim _{k ightarrow infty} a_{n_{k}}=ell.]

Proof

Suppose (limsup _{n ightarrow infty} a_{n}=ell). Then (lim _{n ightarrow infty} s_{n}=ell), where (S_{n}) is defined as in (2.8). For any (varepsilon>0), there exists (N in mathbb{N}) such that

[ell-varepsilon

This implies (s_{N}=sup left{a_{n}: n geq N ight}

[a_{n}

Moreover, for (varepsilon=1), there exists (N_{1} in mathbb{N}) such that

[ell-1

Thus, there exists (n_{1} in mathbb{N}) such that

[ell-1

For (varepsilon=frac{1}{2}), there exists (N_{2} in mathbb{N}) and (N_{2}>n_{1}) such that

[ell-frac{1}{2}

Thus, there exists (n_{2}>n_{1}) such that

[ell-frac{1}{2}

In this way, we can construct a strictly increasing sequence (left{n_{k} ight}) of positive integers such that

[ell-frac{1}{k}

Therefore, (lim _{k ightarrow infty} a_{n_{k}}=ell).

We now prove the converse. Given any (varepsilon>0), there exists (N in mathbb{N}) such that

[a_{n}

for all (n geq M) and (k geq N). Let any (m geq N), we have

[s_{m}=sup left{a_{k}: k geq m ight} leq ell+varepsilon.]

By Lemma 2.1.8, (n_{m} geq m), so we also have

[s_{m}=sup left{a_{k}: k geq m ight} geq a_{n_{m}}>ell-varepsilon.]

Therefore, (lim _{m ightarrow infty} s_{m}=limsup _{m ightarrow infty} a_{n}=ell). (square)

The following result is proved in a similar way.

Theorem (PageIndex{5})

Let (left{a_{n} ight}) be a sequence and (ell in mathbb{R}). The following are equivalent:

1. (liminf _{n ightarrow infty} a_{n}=ell).
2. For any (varepsilon>0), there exists (N in mathbb{N}) such that

[a_{n}>ell-varepsilon ext { for all } n geq N,]

and there exists a subsequence of (left{a_{n_{k}} ight}) of (left{a_{n} ight}) such that

[lim _{k ightarrow infty} a_{n_{k}}=ell.]

Proof

Add proof here and it will automatically be hidden

The following corollary follows directly from Theorems 2.5.4 and 2.5.5.

Corollary (PageIndex{6})

Let (left{a_{n} ight}) be a sequence. Then

[lim _{n ightarrow infty} a_{n}=ell ext { if and only if } limsup _{n ightarrow infty} a_{n}=liminf _{n ightarrow infty} a_{n}=ell .]

Proof

Add proof here and it will automatically be hidden

Corollary (PageIndex{7})

Let (left{a_{n} ight}) be a sequence.

1. Suppose (limsup _{n ightarrow infty} a_{n}=ell) and (left{a_{n_{k}} ight}) is a subsequence of (left{a_{n} ight}) with

[lim _{k ightarrow infty} a_{n_{k}}=ell^{prime}.]

Then (ell^{prime} leq ell).

1. Suppose (liminf _{n ightarrow infty} a_{n}=ell) and (left{a_{n_{k}} ight}) is a subsequence of (left{a_{n} ight}) with

[lim _{k ightarrow infty} a_{n_{k}}=ell^{prime}.]

Then (ell^{prime} geq ell).

Proof

We prove only (a) because the proof of (b) is similar. By Theorem 2.5.4 and the definition of limits, for any (varepsilon>0), there exists (N in mathbb{N}) such that

[a_{n}

for all (n geq N) and (k geq N). Since (n_{N} geq N), this implies

[ell^{prime}-varepsilon

Thus, (ell^{prime}

Remark (PageIndex{8}): Subsequential Limit

Let (left{a_{n} ight}) be a bounded sequence. Define

[A=left{x in mathbb{R}: ext { there exists a subsequence }left{a_{n_{k}} ight} ext { with } lim a_{n_{k}}=x ight}.]

Each element of the set (A) called a subsequential limit of the sequence (left{a_{n} ight}). It follows from Theorem 2.5.4, Theorem 2.5.5, and Corollary 2.5.7 that (A eq emptyset) and

[limsup _{n ightarrow infty} a_{n}=max A ext { and } liminf _{n ightarrow infty} a_{n}=min A.]

Theorem (PageIndex{9})

Suppose (left{a_{n} ight}) is a sequence such that (a_{n}>0) for every (n in mathbb{N}) and

[limsup _{n ightarrow infty} frac{a_{n+1}}{a_{n}}=ell<1.]

Then (lim _{n ightarrow infty} a_{n}=0).

Proof

Choose (varepsilon>0) such that (ell+varepsilon<1). Then there exists (N in mathbb{N}) such that

[frac{a_{n+1}}{a_{n}}

Let (q=ell+varepsilon). Then (0

[0

Since (lim _{n ightarrow infty} q^{n-N} a_{N}=0), one has (lim _{n ightarrow infty} a_{n}=0). (square)

By a similar method, we obtain the theorem below.

Theorem (PageIndex{10})

Suppose (left{a_{n} ight}) is a sequence such that (a_{n}>0) for every (n in mathbb{N}) and

[liminf _{n ightarrow infty} frac{a_{n+1}}{a_{n}}=ell>1.]

Then (lim _{n ightarrow infty} a_{n}=infty).

Proof

Add proof here and it will automatically be hidden

Example (PageIndex{1})

Given a real number (alpha), define

[a_{n}=frac{alpha^{n}}{n !}, n in mathbb{N}. onumber]

Solution

When (alpha =0), it is obvious that (lim _{n ightarrow infty} a_{n}=0). Suppose (alpha>0). Then

[limsup _{n ightarrow infty} frac{a_{n+1}}{a_{n}}=lim _{n ightarrow infty} frac{alpha}{n+1}=0<1. onumber]

Thus, (lim _{n ightarrow infty} a_{n}=0). In the general case, we can also show that (lim _{n ightarrow infty} a_{n}=0) by considering (lim _{n ightarrow infty}left|a_{n} ight|) and using Exercise 2.1.3.

All sequences in this set of exercises are assumed to be in (mathbb{R}).

Exercise (PageIndex{1})

Find (limsup _{n ightarrow infty} a_{n}) and (liminf _{n ightarrow infty} a_{n}) for each sequence.

1. (a_{n}=(-1)^{n}).
2. (a_{n}=sin left(frac{n pi}{2} ight)).
3. (a_{n}=frac{1+(-1)^{n}}{n}).
4. (a_{n}=n sin left(frac{n pi}{2} ight)).

Exercise (PageIndex{2})

For a sequence (left{a_{n} ight}), prove that:

1. (liminf _{n ightarrow infty} a_{n}=infty) if and only if (lim _{n ightarrow infty} a_{n}=infty).
2. (limsup _{n ightarrow infty} a_{n}=-infty) if and only if (lim _{n ightarrow infty} a_{n}=-infty).

Exercise (PageIndex{3})

Let (left{a_{n} ight}) and (left{b_{n} ight}) be bounded sequences. Prove that:

1. (sup _{k geq n}left(a_{n}+b_{n} ight) leq sup _{k geq n} a_{k}+sup _{k geq n} b_{k}).
2. (inf _{k geq n}left(a_{n}+b_{n} ight) geq inf _{k geq n} a_{k}+inf _{k geq n} b_{k}).

Exercise (PageIndex{4})

Let (left{a_{n} ight}) and (left{b_{n} ight}) be bounded sequences.

1. Prove that (limsup _{n ightarrow infty}left(a_{n}+b_{n} ight) leq limsup _{n ightarrow infty} a_{n}+limsup _{n ightarrow infty} b_{n}).
2. Prove that (liminf _{n ightarrow infty}left(a_{n}+b_{n} ight) geq liminf _{n ightarrow infty} a_{n}+liminf _{n ightarrow infty} b_{n}).
3. Find the two counter examples to show that the equalities may not hold in part (a) and part (b).

Exercise (PageIndex{5})

Let (left{a_{n} ight}) be a convergent sequence and let (left{b_{n} ight}) be an arbitrary sequence. Prove that

1. (limsup _{n ightarrow infty}left(a_{n}+b_{n} ight)=limsup _{n ightarrow infty} a_{n}+limsup _{n ightarrow infty} b_{n}=lim _{n ightarrow infty} a_{n}+limsup _{n ightarrow infty} b_{n}).
2. (liminf _{n ightarrow infty}left(a_{n}+b_{n} ight)=liminf _{n ightarrow infty} a_{n}+liminf _{n ightarrow infty} b_{n}=lim _{n ightarrow infty} a_{n}+liminf _{n ightarrow infty} b_{n}).