Prove that every sequence of real numbers (an) has a subsequence (ank) that converges to lim sup

Step 1: Define the lim sup.u000ALet $M u003D u005Climsup_{n u005Cto u005Cinfty} a_n$. By definition, $M$ is the supr

Question

Prove that every sequence of real numbers $\left(a_n\right)$ has a subsequence $\left(a_{n_k}\right)$ that converges to lim sup $n_{n \rightarrow \infty} a_n$. [Hint: If $M=\lim \sup _{n \rightarrow \infty} a_n= \pm \infty$, we must interpret the conclusion loosely; this case is handled in Exercise 25. If $M \neq \pm \infty$, use (*) to choose ( $a_{n_k}$ ) satisfying $\left|a_{n_k}-M\right|<1 / k$, for example.] There is necessarily also a subsequence that converges to liminf ${ }_{n \rightarrow \infty} a_n$. Why?

   Prove that every sequence of real numbers $\left(a_n\right)$ has a subsequence $\left(a_{n_k}\right)$ that converges to lim sup $n_{n \rightarrow \infty} a_n$. [Hint: If $M=\lim \sup _{n \rightarrow \infty} a_n= \pm \infty$, we must interpret the conclusion loosely; this case is handled in Exercise 25. If $M \neq \pm \infty$, use (*) to choose ( $a_{n_k}$ ) satisfying $\left|a_{n_k}-M\right|<1 / k$, for example.] There is necessarily also a subsequence that converges to liminf ${ }_{n \rightarrow \infty} a_n$. Why?

Real analysis

N. L. Carothers 1st Edition

Chapter 1, Problem 27

Instant Answer

Step 1

Let $M = \limsup_{n \to \infty} a_n$. By definition, $M$ is the supremum of the set of all subsequential limits of the sequence $(a_n)$. This means $M$ is the largest value that the subsequences of $(a_n)$ converge to, or it is $\infty$ if no such largest value Show more…

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Prove that every sequence of real numbers $\left(a_n\right)$ has a subsequence $\left(a_{n_k}\right)$ that converges to lim sup $n_{n \rightarrow \infty} a_n$. [Hint: If $M=\lim \sup _{n \rightarrow \infty} a_n= \pm \infty$, we must interpret the conclusion loosely; this case is handled in Exercise 25. If $M \neq \pm \infty$, use (*) to choose ( $a_{n_k}$ ) satisfying $\left|a_{n_k}-M\right|<1 / k$, for example.] There is necessarily also a subsequence that converges to liminf ${ }_{n \rightarrow \infty} a_n$. Why?

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