Max of Sequence a

[happyhim3]

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.

[Amr Morsi]

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.

[Bignose]

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.

[Amr Morsi]

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

[the tree]

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.