This is high school problem only still I don't know where else this question should belong. So posting here,
If for any n greater than or equal to a where a is a natural number, then the least value of a is ?.
This is high school problem only still I don't know where else this question should belong. So posting here,
If for any n greater than or equal to a where a is a natural number, then the least value of a is ?.
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):
Define .
Or in another form ...
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 . 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.
... 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 . 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 . Using parts, the indefinite integral of . Since L'Hospital give that , 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.
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 . 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.