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Topic: Mathematical Real Analysis

- Let (xn) be a bounded sequence ((xn) is not necessarily convergent), and assume that yn → 0. Show that lim n→∞ (xnyn) = 0.

Question1.

All the solution state that there exists M >0 and xn<=M . My question is that why M always be bigger than 0 and Why it is bounded above ? why it is not m<=xn bounded below????

Question. 2.

if the sequence is convergent, then the sequence is bounded. However, lim an=a , is a always be positive ? or it can be converges to negative number?

Question.3

is this convergent sequence bounded below or above? because it has up and down.

above is this bounded or belar ? ----

If you just solve and without answer my confusion, you will get thumb down

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Answer #1

for your question first you need to know the definition of bounded sequence, Defi- A sequence I s nj is bounded if its sangeseal numbers) Isnin-ol = pn Inl=1nlluni < Min. <he Anzi รักปัก ol<a / n>ป Hence for eso 7 an integer In Inoke Knew lim Son Jnes m<xn, we can not derive show that ใกปัก < of you take me on = -m>~) In<-m In 1-2m Jn ol = hanin) = lon) lyn | becomes sonfor m<xn, we cannot delive show that ไปัก oi<e FR you take me sn =) ..m>_), In som In 1-2mJn ol = hinin) =3G) | In). becomes{1,2,...f converges to ! HHHS Converges to in but laring does not converger to either indy 3 Since every convergent Sequence

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