6. Let (xn) be a Cauchy sequence.

6. Let (xn) be a Cauchy sequence. Show that the sequence (x2021n ) converges.7.Let (xn) be a bounded sequence in R. Given N 2 N, we dene the sequences uN andvN as follows:uN := supfxn : n Ng = supfxN; xN+1; xN+2; xN+3; : : :gvN := 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 supn!1xn := limN!1uN = limN!1supfxn : n Ng;lim infn!1xn := limN!1vN = limN!1inffxn : n Ng:Whilst the original sequence (xn) may or may not converge, the lim sup and lim infalways converge (i.e. are always well-dened) if (xn) is bounded.(b) Let’s rst do an example. Let xn = (ô€€€1)n + 1n. Does xn converge? Find supfxn :n 2 Ng and inffxn : n 2 Ng. Given N 2 N, nd the sequences uN, vN, andcompute lim sup xn and lim inf xn. Show that there are subsequences of (xn) whichconverge to lim sup xn and lim inf xn, respectively. Does lim inf xn = sup xn? Whatabout lim inf and inf?(c) Show that, in general, there exists a subsequence (xnk) such that limk!1xnk = lim supn!1xn,and likewise, there is a subsequence (xn`) such that limk!1xn` = lim infn!1xn.Hint: argue similar to Q1 (b) above.(d) Prove that we always have lim infn!1xn lim supn!1xn. Provide an example to showthat the inequality can be strict.