Show that ${\frac{n!}{1\cdot3\cdot5\cdot...\cdot(2n-1)}}$ is monotonic and bounded.

Question

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.

Answer 1

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.