From: Anders Sandberg (asa@nada.kth.se)
Date: Tue Feb 11 2003 - 14:23:33 MST
On Mon, Feb 10, 2003 at 10:33:42PM -0800, Lee Corbin wrote:
> Yes, I know I should re-subscribe to sci.math and all, but
> this is easier---and besides, it may amuse someone here.
>
> I am very sure that for n >= 5, the nth root of n+1
> is less than (2n)/(n+1). But how can it be shown in
> a nice way?
Take the natural log of both objects: (1/n)log n [expression 1] vs.
log(2)+log(n)-log(n+1) [expression 2]. We want to see whether the first
is always smaller than the second. log(n+1) < log(n) + 1/n and log(n) <
log(n+1)-1/(n+1) since log(n) is a convex function (easy to see if you
draw the tangents to the curve). Hence log(n)-log(n+1) < -1/(n+1) and
expression 2 becomes log(2)+log(n)-log(n+1) < log(2) - 1/(n+1) This
means that as n grows it becomes closer to log(2). Meanwhile expression
1 decreases for n larger than e, and approaches zero. Thus expression 1
is smaller for sufficiently large n than expression 2, and this means
that the original statement is true.
I don't know if this is a nice way, but it is straighforward. There is
probably far cooler proofs.
-- ----------------------------------------------------------------------- Anders Sandberg Towards Ascension! asa@nada.kth.se http://www.nada.kth.se/~asa/ GCS/M/S/O d++ -p+ c++++ !l u+ e++ m++ s+/+ n--- h+/* f+ g+ w++ t+ r+ !y
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