the proof I don't like

gistfile1.txt

Let (a_n)_{n=1}^{\inf} be the sequence defined by a_n := n'th_root(c), for some given 0 < c real.
Show that lim_{n -> \inf} a_n = 1.

Proof:

By Bernoulli's inequality,

  (1 + (c-1)/n)^n >= 1 + n * (c-1)/n = c
  
for n >= |c-1|. Taking n'th roots on both sides, we have

  a_n = n'th_root(c) <= 1 + (c-1)/n    for n >= |c-1|.
  
c' := 1/c is also >0, and since

  1/a_n = 1/(c^(1/n)) = (1/c)^(1/n) = c'^(1/n)
  
by the same argument as before, we get that

   1/a_n <= 1 + (c'-1)/n
=> a_n >= 1/(1 + (c'-1)/n)

And we can now sandwich the sequence via:

  1/(1 + (c'-1)/n) <= a_n <= 1 + (c-1)/n
  for n >= max(|c-1|, |1/c-1|).
  
[At this point you still have to produce a sufficient n for a given
epsilon, but that part is routine and has no surprises.]