Once I show that the sequence is monotonic decreasing using the definition $a_n\geq a_{n+1}$, is it sufficient to say "Since it is a nonnegative decreasing sequence, then its greatest lower bound is zero", or do I have to show $\lim_{n\to\infty}{\frac{n!}{1\cdot3\cdot5\cdot...\cdot(2n-1)}}=0$ using a comparison sequence (in which I am unsure what to compare and squeeze it with). Note: I cannot use the Epsilon-N definition per my professor's instructions.
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1If the sequence is $a_n$, then $a_{n+1}/a_n = (n+1)/(2n+1) < 1$. – Kangyeon Moon Jul 23 '23 at 04:09
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I upvoted PrincessEev's answer because it answers exactly to the question you formulated. If your intended question was more demanding, may be you should edit your post to make it explicit. – Anne Bauval Jul 23 '23 at 05:39
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It is not even clear whether your problem is to show that the sequence is monotonic and bounded (title) or that it converges to $0$ (body). – Anne Bauval Jul 23 '23 at 05:46
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Does this answer your question? Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? especially that answer – Anne Bauval Jul 23 '23 at 05:54
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Hello, thank you for the link. I do not understand how showing $a_n<(0.4)(0.6)^{n-2}$ implies that the sequence converges to zero. Is it an application of the squeeze theorem? – dutch Jul 23 '23 at 06:21
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is it sufficient to say "Since it is a nonnegative decreasing sequence, then its greatest lower bound is zero"
Well, no -- mainly for the reason that you don't know that its greatest lower bound would be zero. (Consider, for instance, $a_n = 1 + \frac 1 n$.)
But yes, a non-negative decreasing sequence must be bounded. In particular, $|a_n| \le a_1$ (or whatever the first index of concern is) for any such sequence.
One should also note that, yes, monotone bounded sequences are necessarily convergent. Of course, this does not make a claim as to what the limit is, so if your goal is to show that your sequence approaches $0$, more work needs to be done.
PrincessEev
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Hello, thank you for your answer. If my understanding is correct, since $0 < a_n < a_1$, and it is a monotonic decreasing sequence, then an infimum must exist? – dutch Jul 23 '23 at 05:59
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Yes, and in particular you should be able to show that the infimum in question is the limit of the sequence as well. – PrincessEev Jul 23 '23 at 07:25
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