Solution:

By calculation,

radical n to the radical (n+1) < radical (n+1) to the radical n

for n = 1 through 6, and

radical n to the radical (n+1) > radical (n+1) to the radical n

for n= 7 and 8.

What happens as n gets larger?

Algebraic manipulation tells us that the question as posed is equivalent to asking whether

[radical (n+1)]ln n   <>?  (radical n) ln (n+1)

Consider the expression (radical n) (ln n).  Our problem is equivalent to asking whether this expression will grow faster if we increase the number under the radical by 1 or increase the number under the natural log by 1.  Rate of growth is equivalent to slope: that is, the derivative.

The derivative of radical n is 1/ 2radicaln, while the derivative of ln n is 1/n.  For n =5 or more, 1/2radicaln is greater than 1/n.  As a result, for n sufficiently large, the slope is greater when we increase the radicaln expression by 1 than when we increase the ln expression by 1.  SO, for all n >=7,

radical n to the radical (n+1) > radical (n+1) to the radical n

Dave Stone