Hello, this is my first thread and first post. I was bored today and came up with this interesting problem that I cannot solve (probably to do with the fact that I'm only in calculus II). We are going through sequences so this is my inspiration. So here it goes. (Note: I'll be using a sub n in place of n being a subscript of a because I don't know how to do that on this forum.)

 

Let a be a sequence with a sub n being the nth term. Let a sub n=n^(n^x)/(n!)^n. Find max. a sub n in terms of x where x satisfies that the limit of a sub n as n approaches infinity equals zero and n is in the natural number set.

 

Have fun.

Differentiate with respect to n and equate to zero. And, then find the nearest natural number. Note that x will be considered a constant. Also, I think that n! will be considered constant as differentiation is over a small variation around n.

Differentiate with respect to n and equate to zero. And, then find the nearest natural number. Note that x will be considered a constant. Also, I think that n! will be considered constant as differentiation is over a small variation around n.

 

If you are going to differentiate with respect to n, you cannot let n! be a constant. n! is the term with the greatest rate of change for large n in that expression.

 

To do it right, you'd use the Gamma function as the extension of the factorial to the reals, and use the derivative of the Gamma function.

You are right Bignose. But, isn't the power here is most significant: (n!)^n. Plus, at large n the sequence turns to zero.

You are right Bignose. But, isn't the power here is most significant: (n!)^n.

The difference in the derviatives of the denominator seem too big to ignore when you take n! to be constant.