n = (1 + d)n > n(n-1)
2
 d2
so
d2 < 2
n-1

and
n1/n = 1 + d < 1 +   æ
 ú
Ö
2
n-1
 
 .
Hence d® 0 as n® ¥ and n1/n® 1 as n ® ¥.

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. We need to prove that there is no large value of n where the value is larger than this and clearly it is impossible to check all values of n. However, using the earlier result, if n ³ 19 then
n1/n < 1 +   æ
 ú
Ö
2
18
 
 = 1+ 1/3 < Ö2

so the maximum occurs within the set where n=1 to 18. It can be checked that the maximum is 31/3.