Stirling’s approximation, but weaker

Show that

\[\lim_{n\to \infty} \frac{e^nn!}{n^n} = \infty.\]

You may use the power series for the exponential function.

Hint

Keep several terms just after \(n^n/n!\) in the series for \(e^n\).

Solution

Fix a positive integer \(k\). Keeping \(k+1\) terms from the exponential series gives

\[\frac{e^nn!}{n^n} \geq 1+\sum_{j=1}^{k}\frac{n^j n!}{(n+j)!}. \tag{1}\label{eq:factorial-tail}\]

Each ratio \(n^j n!/(n+j)!\) is a product of \(j\) factors \(n/(n+r)\), each tending to \(1\). Thus the right-hand side of \(\eqref{eq:factorial-tail}\) tends to \(k+1\).

Given any bound \(M\), choose \(k\) with \(k+1>M\). Then \(\eqref{eq:factorial-tail}\) exceeds \(M\) for all sufficiently large \(n\), proving the limit.