## Monotone Decreasing

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?
 PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract

Recognitions:
Homework Help
Science Advisor
 Quote by Hammie 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?
 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?

Recognitions:
Homework Help
Science Advisor

## Monotone Decreasing

That would be it.

 Similar discussions for: Monotone Decreasing Thread Forum Replies Calculus & Beyond Homework 2 Calculus & Beyond Homework 1 Calculus 8 Calculus & Beyond Homework 3 Linear & Abstract Algebra 2