Prove "Monotone Decreasing" of (1 + 1/x)^(x+1)

[Hammie]

Here's another one I'm doing just for the fun of it..

"prove that (1 + 1/x) ^ (x + 1) is monotone decreasing"

Okie Dokie..

If it just said show it, I'd be happy. Just plug in n=2, 3, 4.. and it is easy enough to observe that each term is decreasing.

But to prove it is monotone decreasing I must show that a(n) must be greater than a(n-1), that a(n)/a(n-1) < 1, at least for all large n.

What I have so far:

a(n)/a(n-1) = (1 + 1/n)((n^2-1)/n^2)^n

Or (1 + 1/n)(1 - 1/n^2)^n

What's up with this? How can I prove that this ratio is less than one? If the answer is obvious, it just seems to elude me..

This is early on in an old advanced calculus text. It hasn't even begun to talk about derivatives at this point.. I feel like I'm being asked to perform brain surgery with bone knives and bear skins..




any suggestions or hints?

 

[shmoe]

Or (1 + 1/n)(1 - 1/n^2)^n

At this point it would be nice to have a common power to simplify things. Can you bound (1+1/n) from above by something like (1+x)^n? What will work for x?

 

[Hammie]

OK..

Don't know if this is what you were alluding to, but how about-

Use Bernouli's inequality, kind of backwards.

(1 + 1/n^2)^n > (1 + 1/n). simply reverse: (1+1/n) < (1 + 1/n^2)^n.

Therefore,

(1+1/n)(1 - 1/n^2)^n < (1-1/n^2)^n(1 + 1/n^2)^n = (1 - 1/n^4)^n

which by examination is less than one for all n.

Therefore a(n)/a(n-1) is less than one for all n. a(n) < a(n-1), therefore it is montone decreasing.

Is this valid?

 

[shmoe]

That would be it.