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Prove that if \(p=\lim x_{n}\) in \(E^{1}\), then $$\underline{\lim }\left(x_{n}+y_{n}\right)=p+\underline{\lim } y_{n} ;$$ similarly for \(\bar{L}\).

Short Answer

Expert verified
\(\underline{\lim} (x_n + y_n) = p + \underline{\lim} y_n;\ \overline{\lim} (x_n + y_n) = p + \overline{\lim} y_n.\)

Step by step solution

01

Understanding the Problem

We need to prove that for the sequence of real numbers, if the limit of \(x_n\) exists and is \(p\), then the limit inferior of the sequence \((x_n + y_n)\) is equal to \(p + \underline{\lim} y_n\), and similarly for the limit superior.
02

Recall Definitions

Recall that \( \underline{\lim} a_n = \lim_{n \to \infty} \inf_{k \geq n} a_k \) and \( \overline{\lim} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_k \). These are the limit inferior and limit superior of a sequence.
03

Apply Limits and Properties for \(x_n\)

Given \(p = \lim x_n\), this implies \(x_n \to p\) as \(n \to \infty\). This means that for any \(\epsilon > 0\), there exists some \(N\) such that for all \(n > N\), \(|x_n - p| < \epsilon\).
04

Apply Limits to Combined Sequence

Consider the sequence \(x_n + y_n\). We know that \(\lim x_n = p\), so for \(n\) large enough, \(x_n + y_n\) behaves similar to \(p + y_n\).
05

Prove Limit Inferior Identity

For the limit inferior, we consider \(\underline{\lim} (x_n + y_n) = \lim_{n \to \infty} \inf_{k \ge n} (x_k + y_k)\). Because \(x_n\) approaches \(p\), we have \( (p + \inf_{k \ge n} y_k ) \) for large \(n\), which implies \( \underline{\lim} (x_n + y_n) = p + \underline{\lim} y_n\).
06

Prove Limit Superior Identity

Similarly, for the limit superior, we consider \(\overline{\lim} (x_n + y_n) = \lim_{n \to \infty} \sup_{k \ge n} (x_k + y_k)\). Because \(x_n\) approaches \(p\), this turns into \(p + \sup_{k \ge n} y_k\), leading to \(\overline{\lim} (x_n + y_n) = p + \overline{\lim} y_n\).
07

Conclusion of Proof

We have shown that both identities for limit inferior and limit superior hold: \( \underline{\lim} (x_n + y_n) = p + \underline{\lim} y_n \) and \( \overline{\lim} (x_n + y_n) = p + \overline{\lim} y_n \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Superior
The limit superior (often denoted as \( \overline{\lim} a_n \)) of a sequence is essentially a way to capture the largest value that the sequence approaches as \( n \to \infty \). This isn't necessarily the maximum value of the entire sequence, but rather the supremum (least upper bound) of all possible accumulation points.Think of it as the ceiling, or highest level, that the sequence keeps approaching, even if just for fleetingly short intervals. It is defined mathematically as:\[ \overline{\lim} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_k \]Here is how it works:
  • The expression \( \sup_{k \geq n} a_k \) calculates the smallest number that is greater than or equal to every \( a_k \) for \( k \geq n \).
  • Taking the limit as \( n \rightarrow \infty \), we find the smallest ceiling towards which the tail of the sequence grows.
Understanding limit superior helps us analyze sequences where they might not have a set point to converge to, but they still exhibit bounded behavior closer to infinity.
Limit Inferior
Limit inferior (denoted as \( \underline{\lim} a_n \)) provides us with a way to capture the smallest value that a sequence approaches as \( n \to \infty \). Just like limit superior, it's not necessarily the smallest element in the sequence, but rather the greatest lower bound of all the sequence's possible limit points.The formal definition is:\[ \underline{\lim} a_n = \lim_{n \to \infty} \inf_{k \geq n} a_k \]This can be explained in an approachable way:
  • \( \inf_{k \geq n} a_k \) looks at the largest number \( a_k \) for \( k \geq n \), below which all subsequent terms \( a_k \) fall.
  • By taking the limit as \( n \rightarrow \infty \), we find the highest floor that the tail end of the sequence can be said to consistently reach.
Grasping the limit inferior is crucial when encountering sequences that do not settle at a point but rather waver, understanding the lowest limits of their oscillating patterns.
Convergent Sequences
A convergent sequence is one of the simplest concepts in mathematical analysis, yet incredibly important. A sequence \( \{x_n\} \) is said to converge if it approaches a single finite number \( p \) as \( n \to \infty \). In other words, the further along in the sequence you go, the closer the terms get to \( p \).This can be expressed formally:For every \( \epsilon > 0 \), there exists an \( N \) such that for all \( n > N \), \( |x_n - p| < \epsilon \).Breaking this down, we get:
  • "For every \( \epsilon > 0\)" means no matter how small a margin of error we choose, the sequence will eventually show a consistent trend toward \( p \).
  • Buying an "N" implies that beyond some point in the sequence \( N \), all terms of the sequence are inside the margin we've set.
Convergent sequences show us predictability and stability, such that knowing the limit means we understand the sequence's long-term behavior.

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Most popular questions from this chapter

Prove in \(E^{1}\) that (i) \(\sum_{k=1}^{n} k=\frac{1}{2} n(n+1) ;\) (ii) \(\sum_{k=1}^{n} k^{2}=\frac{1}{6} n(n+1)(2 n+1) ;\) (iii) \(\sum_{k=1}^{n} k^{3}=\frac{1}{4} n^{2}(n+1)^{2}\); (iv) \(\sum_{k=1}^{n} k^{4}=\frac{1}{30} n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)\).

Prove the principle of nested intervals: If \(\left[a_{n}, b_{n}\right]\) are closed intervals in a complete ordered field \(F\), with $$ \left[a_{n}, b_{n}\right] \supseteq\left[a_{n+1}, b_{n+1}\right], \quad n=1,2, \ldots $$ then $$ \bigcap_{n=1}^{\infty}\left[a_{n}, b_{n}\right] \neq \emptyset $$ [Hint: Let $$ A=\left\\{a_{1}, a_{2}, \ldots, a_{n}, \ldots\right\\} $$ Show that \(A\) is bounded above by each \(b_{n}\). Let \(p=\sup A\). (Does it exist?) Show that $$ (\forall n) \quad a_{n} \leq p \leq b_{n} $$ i.e., $$ \left.p \in\left[a_{n}, b_{n}\right] \cdot\right] $$

Prove that (i) \(\overline{\lim }\left(-x_{n}\right)=-\underline{\lim } x_{n}\) and (ii) \(\overline{\lim }\left(a x_{n}\right)=a \cdot \overline{\lim } x_{n}\) if \(0 \leq a<+\infty\).

In an ordered field \(F,\) let \(\emptyset \neq A \subset F\). Let \(c \in F\) and let \(c A\) denote the set of all products \(c x(x \in A) ;\) i.e., $$ c A=\\{c x \mid x \in A\\} $$Prove that (i) if \(c \geq 0\), then $$ \sup (c A)=c \cdot \sup A \text { and } \inf (c A)=c \cdot \inf A $$ (ii) if \(c<0\), then $$ \sup (c A)=c \cdot \inf A \text { and } \inf (c A)=c \cdot \sup A $$ In both cases, assume that the right-side sup \(A\) (respectively, inf \(A\) ) exists.

Prove that if \(\left\\{x_{n}\right\\}\) and \(\left\\{y_{n}\right\\}\) are bounded in \(E^{1}\), then \(\overline{\lim } x_{n}+\overline{\lim } y_{n} \geq \overline{\lim }\left(x_{n}+y_{n}\right) \geq \overline{\lim } x_{n}+\underline{\lim } y_{n}\) \(\geq \underline{\lim }\left(x_{n}+y_{n}\right) \geq \underline{\lim } x_{n}+\underline{\lim } y_{n}\).

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