What is bounded and unbounded sequence?

What is bounded and unbounded sequence?

A sequence an is bounded below if there exists a real number M such that. M≤an. for all positive integers n. A sequence an is a bounded sequence if it is bounded above and bounded below. If a sequence is not bounded, it is an unbounded sequence.

Is it true that every bounded sequence is convergent?

Every bounded sequence is NOT necessarily convergent.

Do bounded sequences converge?

If a sequence an converges, then it is bounded. Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence (−1)n is bounded, but the sequence diverges because the sequence oscillates between 1 and −1 and never approaches a finite number.

Do all bounded sequences have limits?

If a sequence is bounded there is the possibility that is has a limit, though this will not always be the case. If it does have a limit, the bound on the sequence also bounds the limit, but there is a catch which you must be careful of. Theorem giving bounds on limits.

Can a bounded sequence be divergent?

As far as I know a bounded sequence can either be convergent or finitely oscillating, it cannot be divergent since it cannot diverge to infinity being a bounded sequence.

Is a bounded sequence always convergent?

How do you prove that an infinite sequence is bounded?

An infinite sequence can be proved to be bounded if we can prove that the sequence is convergent. This is because convergence means approximating to a finite value, called the sum. In fact, it can be proved that every convergent sequence is bounded.

What is the upper bound for a monotone decreasing sequence?

Also note a 1 = 1 < 3 / 2. If a n < 3 / 2 then a n + 1 ≤ 1 + a n / 3 < 1 + 1 / 2 = 3 / 2 as well. So the entire sequence is bounded above by 3 / 2 by induction. This may be informal, but just showing that the sequence is monotone decreasing would mean the first term is the upper bound for the sequence.

How do you prove something without using the geometric series?

Now use what you know about geometric series. If you want a direct proof that doesn’t use geometric series outright, note that if you consider the terms in a n shifted forward one position and then compare to the terms in a n + 1, you get a n + 1 ≤ 1 + a n / 3. Also note a 1 = 1 < 3 / 2.