Hey, I'm stuck on this question and desperately need help.
Question
Suppose that the sequence is convergent, and assume that the sequence is bounded. Prove that the sequence defined by
is convergent and find its limit. (n belongs to the set of natural numbers)
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Please help! I tried using the theorem that if a sequence is convergent, then it is bounded (for ) but get anywhere. It's quite clear to me that the limit is but I don't know how to get there.
Ok, thanks for the reply. Here's what I've done:
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Let is a natural number , and let be an upper bound of the set.
Let is a natural number , and let be an upper bound of the set.
Then, for all
But as for all and y for all , it follows that
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But I don't know how to prove the sequence is convergent before finding the limit as the question asks.
Could I perhaps show the sequence is monotone and bounded and therefore convergent? I've tried, but couldn't prove it.
Also, what if and ? We made a restriction but surely this doesn't cover all possibilities.
Sorry if I'm asking silly questions, but I really need help with this.
If we do not make this restrictions then the denominator can be zero and the sequences are not well-defined. It is absolutely necessary to make some sort of restriction.
Just use the fact that is is bounded then . Thus, if a sequence is convergent (therefore bounded) we have .
This means,
Because is bounded and is bounded.
Ok, thank you.
In the seminars, we haven't been told that if the sequence is bounded. As obvious as it is.
So do you think it would be more 'rigorous' if I said that since is bounded, there exists an such that
Then I can show that and . Since , then by the sandwich theorem,
Is that fine, in your opinion, or unnecessary?
Thanks.