From: Hal Finney (hal@finney.org)
Date: Tue Feb 11 2003 - 16:31:33 MST
Lee Corbin writes:
> 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?
As n approaches infinity, the nth root of n+1 approaches 1, while
(2n)/(n+1) approaches 2. So obviously this will be true for some
sufficiently large n. You can establish it manually for 5. The
remaining piece would be to show that (n+1)**(1/n) is a decreasing
function for numbers of this size, which I think you could show by
taking the derivative.
My calculus is a little rusty so I will take the natural log of this
expression instead, which is ln(n+1)/n, and I will show that it is
decreasing (I remember how to take the derivative of fractions but not
of exponentials). The derivative of this is (n/(n+1) - ln(n+1))/n^2,
which I claim is negative. For it to be negative ln(n+1) > n/n+1,
which will be true if ln(n+1) > 1 since n/n+1 < 1. ln(n+1) > 1 if n+1 > e,
hencer n > 2. So at least for values n > 2, the nth root of n+1 is a
decreasing function and the whole thing follows.
Hal
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