OK, here goes.
Exercise: Show that \[\frac{n}{{\sqrt[n]{{n!}}}} < \left( {1 + \frac{1}{n}} \right)^n \] for all natural numbers n.
Solution: We use without proof the following inequalities:
(1 + \frac{1}{k})^k \leq e \leq (1 + \frac{1}{k})^{k + 1} \right {(*)}
(where e is Euler's number) for all natural numbers k. A proof may be found in many calculus and real analysis books. (I couldn't find a good online reference for these inequalities. Does anyone know of one?)
For k = 1, 2, ..., n - 1, multiply together the inequalities on the left side of (*):\prod_{k=1}^{n-1}(1 + \frac{1}{k})^{k} \leq e^{n -1}The left side of this inequality equals \frac{n^n}{n!} (To see this, rewrite (1 + \frac{1}{k}) as
\frac{k + 1}{k}, simplify the product to \[\frac{n^{n-1}}{(n-1)!} and multiply by \frac{n}{n}).
Therefore\frac{n^n}{n!} \leq e^{n-1}or,\[\frac{n}{{\sqrt[n]{{n!}}}} \leq e^{\frac{n-1}{n}} = e^{1-\frac{1}{n}}Now, in (*), raise the right inequality to the power 1 - \frac{1}{n}:e^{1-\frac{1}{n}} \leq (1 + \frac{1}{n})^{(n+1)(1-\frac{1}{n})} = (1 + \frac{1}{n})^{n-\frac{1}{n}}, which is strictly less than (1 + <br />
<br />
\frac{1}{n})^{n}.
Therefore, \[\frac{n}{{\sqrt[n]{{n!}}}} \leq e^{1-\frac{1}{n}} < (1 + \frac{1}{n})^{n}, as required.
