Limit Inferior and Limit Superior We define the limit inferior and the limit superior of a sequence as follows: lim inf an = lim inf{ak | k≥n} = 0+ (n→∞) lim sup an = lim sup{ak | k≥n} = ∞ (n→∞) Let {an} be a bounded sequence. Show that lim inf an and lim sup an exist and are in â„. (n→∞) Let {an} be an unbounded sequence. Show that either lim inf an = -∞ or lim sup an = ∞ (or possibly both). Let {an}, {bn} be two sequences. Show that lim inf (an + bn) ≤ lim inf an + lim sup bn ≤ lim sup an + lim sup bn (n→∞) Furthermore, find a pair of sequences for which the second inequality is strict.
The Correct Answer and Explanation is:
Definitions:
- Limit Inferior (lim inf) of a sequence {an}\{a_n\}{an}: lim infn→∞an=limn→∞(inf{ak∣k≥n})\liminf_{n \to \infty} a_n = \lim_{n \to \infty} \left( \inf \{a_k | k \geq n\} \right)n→∞liminfan=n→∞lim(inf{ak∣k≥n}) This is the greatest lower bound of the tail of the sequence as n→∞n \to \inftyn→∞. Intuitively, it represents the lowest value the sequence “settles” on in the long run.
- Limit Superior (lim sup) of a sequence {an}\{a_n\}{an}: lim supn→∞an=limn→∞(sup{ak∣k≥n})\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \left( \sup \{a_k | k \geq n\} \right)n→∞limsupan=n→∞lim(sup{ak∣k≥n}) This is the least upper bound of the tail of the sequence as n→∞n \to \inftyn→∞. Intuitively, it represents the highest value the sequence “settles” on in the long run.
Bounded Sequence:
Let {an}\{a_n\}{an} be a bounded sequence. This means that there exist constants MMM and mmm such that:m≤an≤Mfor alln.m \leq a_n \leq M \quad \text{for all} \quad n.m≤an≤Mfor alln.
We need to show that both lim infan\liminf a_nliminfan and lim supan\limsup a_nlimsupan exist and are real numbers.
Existence:
- For each nnn, we can consider the set {ak∣k≥n}\{ a_k \mid k \geq n \}{ak∣k≥n}, which is a subset of the real numbers and is bounded because ana_nan is bounded.
- The infimum (greatest lower bound) and supremum (least upper bound) of this set will be finite because the set of real numbers is complete and the sequence is bounded.
- Since the infimum and supremum are non-decreasing and non-increasing respectively as nnn increases, the limits: limn→∞inf{ak∣k≥n}andlimn→∞sup{ak∣k≥n}\lim_{n \to \infty} \inf \{a_k \mid k \geq n\} \quad \text{and} \quad \lim_{n \to \infty} \sup \{a_k \mid k \geq n\}n→∞liminf{ak∣k≥n}andn→∞limsup{ak∣k≥n} exist by the monotone convergence theorem. Therefore, both the limit inferior and limit superior are well-defined and finite, and they are real numbers.
Unbounded Sequence:
Let {an}\{a_n\}{an} be an unbounded sequence. This means that the sequence does not remain within any finite bounds, so either {an}\{a_n\}{an} tends to infinity or negative infinity at some points.
Proof:
- If {an}\{a_n\}{an} is unbounded above, this implies there exists a subsequence where an→+∞a_n \to +\inftyan→+∞, hence: lim supn→∞an=∞.\limsup_{n \to \infty} a_n = \infty.n→∞limsupan=∞.
- If {an}\{a_n\}{an} is unbounded below, this implies there exists a subsequence where an→−∞a_n \to -\inftyan→−∞, hence: lim infn→∞an=−∞.\liminf_{n \to \infty} a_n = -\infty.n→∞liminfan=−∞.
Thus, if a sequence is unbounded, we can have either lim infan=−∞\liminf a_n = -\inftyliminfan=−∞ or lim supan=∞\limsup a_n = \inftylimsupan=∞, or both.
Inequalities involving lim inf and lim sup:
Now, consider two sequences {an}\{a_n\}{an} and {bn}\{b_n\}{bn}. We want to show the following inequalities:lim infn→∞(an+bn)≤lim infn→∞an+lim supn→∞bn≤lim supn→∞(an+bn).\liminf_{n \to \infty} (a_n + b_n) \leq \liminf_{n \to \infty} a_n + \limsup_{n \to \infty} b_n \leq \limsup_{n \to \infty} (a_n + b_n).n→∞liminf(an+bn)≤n→∞liminfan+n→∞limsupbn≤n→∞limsup(an+bn).
To show this:
- First Inequality: lim inf(an+bn)≤lim infan+lim supbn\liminf (a_n + b_n) \leq \liminf a_n + \limsup b_nliminf(an+bn)≤liminfan+limsupbn The sequence an+bna_n + b_nan+bn is formed by adding the sequences ana_nan and bnb_nbn. By the definition of lim inf, the limit inferior of the sum is at most the sum of the limit inferior of ana_nan and the limit superior of bnb_nbn. The intuition is that the infimum of a sum is less than or equal to the sum of the infimum of one sequence and the supremum of the other.
- Second Inequality: lim sup(an+bn)≥lim supan+lim supbn\limsup (a_n + b_n) \geq \limsup a_n + \limsup b_nlimsup(an+bn)≥limsupan+limsupbn Similarly, the supremum of a sum is at least the sum of the suprema of the individual sequences.
Finding Strict Inequality:
Consider the following example:
- Let an=(−1)na_n = (-1)^nan=(−1)n and bn=(−1)n+1b_n = (-1)^{n+1}bn=(−1)n+1.
- For an=(−1)na_n = (-1)^nan=(−1)n, the lim inf and lim sup are: lim infan=−1,lim supan=1.\liminf a_n = -1, \quad \limsup a_n = 1.liminfan=−1,limsupan=1.
- For bn=(−1)n+1b_n = (-1)^{n+1}bn=(−1)n+1, the lim inf and lim sup are: lim infbn=−1,lim supbn=1.\liminf b_n = -1, \quad \limsup b_n = 1.liminfbn=−1,limsupbn=1.
- Now, consider the sum: an+bn=(−1)n+(−1)n+1.a_n + b_n = (-1)^n + (-1)^{n+1}.an+bn=(−1)n+(−1)n+1. This sequence alternates between 0 and -2. Thus: lim inf(an+bn)=−2,lim sup(an+bn)=0.\liminf (a_n + b_n) = -2, \quad \limsup (a_n + b_n) = 0.liminf(an+bn)=−2,limsup(an+bn)=0.
In this case:lim inf(an+bn)=−2,lim infan+lim supbn=−1+1=0.\liminf (a_n + b_n) = -2, \quad \liminf a_n + \limsup b_n = -1 + 1 = 0.liminf(an+bn)=−2,liminfan+limsupbn=−1+1=0.
Here, the second inequality is strict:lim inf(an+bn)<lim infan+lim supbn.\liminf (a_n + b_n) < \liminf a_n + \limsup b_n.liminf(an+bn)<liminfan+limsupbn.
This demonstrates that the second inequality can indeed be strict in certain cases.
