Prove that if {an} is a bounded sequence (not necessarily convergent) and {bn} converges to 0, then anbn 0. Use this result to prove that sin n COS n lim 818 n = lim 818 n = 0. ->
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- Show that the sequence {Sn} give in general terms is convergent.Suppose that {xn} is a bounded sequence such that all convergent subsequences converge to the same numbera. Prove that {xn} must converge to a.Let {an} be a monotonic sequence such that an ≤ 1. Discuss the convergence of {an}. When {an} converges, what can you conclude about its limit?
- 2) Show that the sequence defined recursively by 3 4 An aj = 5 аn+1 satisfies 2 < an < 5 and is decreasing. Show that the sequence is convergent and find its limit.Determine whether {an} is convergent(6) Let {an} be the sequence defined by an+1 = √6an — 2, a₁ = 1. Show that {an} is convergent and find the limit.
- Consider the sequence 6" an 2+7" * Determine if the sequence converges or diverges. If it converges, what is the limit?Find 2x xp (if it converges)Determine whether the sequence converges or diverges and if it converges find its limit.a_n= (1*3*5* * *(2n-1))/n! Also, see attached picture of the problem, it is better. Thank you.