= Solution
By part (e) and the <comparison test for series>, $\sum u_n$ converges, since the <P-series> $\sum n^{-2}$ converges. Put $S=\sum_{n=1}^{\infty}u_n$. Part (a) now proves
$$
\boxed{a_n\longrightarrow e^{-1+S}\in(0,\infty).}
$$
Part (d) also shows that $(a_n)$ increases towards this finite positive <limit>. This proves <convergence of the Stirling sequence> without presupposing its limiting constant.
For identification of that constant, the independently known <Stirling's formula> $n!\sim\sqrt{2\pi n}(n/e)^n$ gives
$$
\boxed{\lim_{n\to\infty}a_n=\frac1{\sqrt{2\pi}}.}
$$
The comparison argument above supplies the convergence proof requested by the preceding parts; the <Stirling's formula> identification is an additional evaluation.
Back to article page
