Mathematics Homework Help

6. Let (xn) be a Cauchy sequence. Show that the sequence (x2021

n ) converges.

7.Let (xn) be a bounded sequence in R. Given N 2 N, we dene the sequences uN and

vN as follows:

uN := supfxn : n Ng = supfxN; xN+1; xN+2; xN+3; : : :g

vN := inffxn : n Ng = inffxN; xN+1; xN+2; xN+3; : : :g:

(a) Prove that the sequences fuNg and fvNg converge as N ! 1.

We introduce the following notations:

lim sup

n!1

xn := lim

N!1

uN = lim

N!1

supfxn : n Ng;

lim inf

n!1

xn := lim

N!1

vN = lim

N!1

inffxn : n Ng:

Whilst the original sequence (xn) may or may not converge, the lim sup and lim inf

always converge (i.e. are always well-dened) if (xn) is bounded.

(b) Let’s rst do an example. Let xn = ( 1)n + 1

n. Does xn converge? Find supfxn :

n 2 Ng and inffxn : n 2 Ng. Given N 2 N, nd the sequences uN, vN, and

compute lim sup xn and lim inf xn. Show that there are subsequences of (xn) which

converge to lim sup xn and lim inf xn, respectively. Does lim inf xn = sup xn? What

about lim inf and inf?

(c) Show that, in general, there exists a subsequence (xnk) such that lim

k!1

xnk = lim sup

n!1

xn,

and likewise, there is a subsequence (xn`) such that lim

k!1

xn` = lim inf

n!1

xn.

Hint: argue similar to Q1 (b) above.

(d) Prove that we always have lim inf

n!1

xn lim sup

n!1

xn. Provide an example to show

that the inequality can be strict.