Congratulations to Curt from Reigate College for cracking this tough
nut problem. Here is Curt's solution:
If n = n1/n, I first assert that for an n > 1, then
n1/n is larger than 1
If n > 1 then, raising both sides to the power of 1/n yields
n1/n > 11/n. So n1/n is larger than 1 no
matter what the value of 1/n. Therefore, one could make n1/n=1+R. As n1/n is always larger than 1, R is always a
positive real number. Now, raising both sides to the power n one
obtains:
n = 1 + nR +
12
n(n-1)R2 +¼.
(Incidentally, n > nR, therefore R < 1 and so n1/n < 2)
As n is an integer larger than one, at least the third term of
expansion must exist. It is clear that:
n
> 1/2n(n-1)R2
[(2n)/(n(n-1))]
> R2
Ö{[2/(n-1)]}
> R
.
So R® 0 as n® ¥ and
n1/n = 1 + R < 1 +
æ ú
Ö
2n-1
.
So n1/n® 1 as n® ¥.
As a corollary, I will show that
______ Ö2/(n-1)
grows smaller for
successive values of n > 1. Assume that for some positive B,
______ Ö2/(n-1)
<
________ Ö2/(n+B-1)
. Now if this is to be true
1/(n-1) < 1/(n+b-1), therefore n-1 > n+B-1, therefore B < 0,
contrary to our conditions. Therefore the assertion that for some
positive B the value of the function increases is fallacious. This
will prove useful later.
For n = 1, 2, etc. we have n = 1, 21/2, 31/3, 41/4=21/2 ... and we see that the values increase to a
maximum of 31/3 and then start to decrease.
Now to prove that 3 is the value of n pertaining to the maximum of
this discrete function, I will find an n such that
1+
______ Ö2/(n-1)
is less that the cube root of 3. From that point
onwards, it is known that for successive values
______ Ö2/(n-1)
decreases, therefore beyond this point the value of n1/n
decreases; more importantly, is less than 31/3. For n=19
this holds true, as it will for subsequent values. The ïn-between"
values have been checked and are less than 31/3.
If n ³ 19 then