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Thread: Find minimum natural value of n such that (n/2)^n > n! > (n/3)!

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    Find minimum natural value of n such that (n/2)^n > n! > (n/3)!

    This is high school problem only still I don't know where else this question should belong. So posting here,
    If \left( \frac{n}{2} \right)^n > n! >\left( \frac{n}{3} \right)^n<br />
for any n greater than or equal to a where a is a natural number, then the least value of a is ?.
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    Re: Find minimum natural value of n such that (n/2)^n > n! > (n/3)!

    Have you even tried listing a table for the first 10 or so values of $n$ to see what the answer might be?
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    Re: Find minimum natural value of n such that (n/2)^n > n! > (n/3)!

    Quote Originally Posted by romsek View Post
    Have you even tried listing a table for the first 10 or so values of $n$ to see what the answer might be?
    A great little problem to play with in excel.

    I would have wagered that n! would easily outstrip them both ... and would have lost. It's an interesting fact that this inequality holds eventually - wonder where it came up ... besides as a math problem?

    I chased it a little more. It reforms into (using monotonicity of log):

     \log(1/2) < \frac{1}{n}\sum_{k=1}^{n}{\log(k/n)} < \log(1/3)

    Define x_n \doteqdot \frac{1}{n}\sum_{k=1}^{n}{\log(k/n)}.

    Or in another form ...

    x_n = \sum_{k=1}^{n}{\log \left[ (k/n)^\frac{1}{n}\right]}

    It's not a series - each would-be "partial sum" is different - changes with n. So we need monotonicity for the bound to help us; a little work shows that the sequence is, indeed, monotone decreasing. So this sequence converges to some negative value greater than \log(1/3) \approx -0.477. I have no idea how to identify the number it converges to ... it's probably irrational and involves e and pi somehow.

    Value is about -0.4324 at n = 1000 ... fairly slow convergence which you would expect from a the log(). I haven't met a sequence like this before. Hmm ... strange little sequence.
    Last edited by mgeile; Apr 2nd 2017 at 02:17 PM.
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    Re: Find minimum natural value of n such that (n/2)^n > n! > (n/3)!

    ... okay ... a mistake on the calculation as I used the wrong function in excel ... log base 10 instead of base e. I can't edit the post as it timed out ... so I'll just state the upgrades. The bounds on the sum are about -0.643 < x_n < -1.099. Since -1 is on this interval, we might suspect that as a possible point of convergence ... but let's see if we can prove it.

    This sum that makes the sequence in the previous post (first version) does look like a common sequence - the partition of log(x) on [0,1] with uniform partitions of width 1/n.

    So the convergence should be to the value  \int_{0}^{1}{\ln(x) dx}. Using parts, the indefinite integral of \ln(x) = x \ln(x) - x. Since L'Hospital give that \lim_{x \rightarrow 0}{(x \ln(x))} = 0, we indeed have that the series converges to -1.

    So, the original problems was not difficult ... given a calculator or other suitable electronic tool. I'm still wondering how the problem's author decided on the stated form - working with the factorial. It probably came about when Euler was playing around one day - when people actually thought about the partitions. Wouldn't it be interesting to know the history of some of these little problems that have been passed about for years.
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    Re: Find minimum natural value of n such that (n/2)^n > n! > (n/3)!

    After mulling on this for awhile, I get back to the primary point of the problem is that there is an "a" such that for all n \ge a, \left( \frac{n}{3} \right)^n < n! < \left( \frac{n}{2} \right)^n. The sequence, discussed above, and it's convergence is the mechanism that you can use to prove the result for all n > a. I'm pretty confident that whomever discovered this inequality was looking for approximations to n! to open up an analysis path.
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