Fall 2013 - Problem 8

Solution.

1. Let $f_k\p{x} = \p{1 - \frac{x}{k}}^k \chi_{\br{0,k}}\p{x}$. By AM-GM,

   $$\p{1 \cdot \p{1 - \frac{x}{k}}^k}^{1/\p{k+1}} \leq \frac{1}{k+1}\p{1 + k \cdot \p{1 - \frac{x}{k}}} = 1 - \frac{x}{k+1},$$

   so for $x \in \br{0, k}$,

   $$\begin{aligned} f_k\p{x} - f_{k+1}\p{x} &= \p{1 - \frac{x}{k}}^k - \p{1 - \frac{x}{k+1}}^{k+1} \\ &\leq 1 - \frac{x}{k+1} - \p{1 - \frac{x}{k+1}}^{k+1}. \end{aligned}$$

   Since $1 - \frac{x}{k+1} \in \br{0, 1}$, we see $f_k\p{x} \leq f_{k+1}\p{x}$ on $\br{0, k}$. On $\poc{k, k+1}$, $f_k\p{x} = 0$ but $f_{k+1}\p{x} \geq 0$, and so $f_k \leq f_k$ everywhere. $f_k\p{x} \to e^{-x}$, so by the monotone convergence theorem,

   $$\lim_{k\to\infty} \int_0^k x^n\p{1 - \frac{x}{k}}^k \,\diff{x} = \int_0^\infty \lim_{k\to\infty} x^nf_k\p{x} \,\diff{x} = \int_0^\infty x^ne^{-x} \,\diff{x} = \Gamma\p{n+1} = n!.$$

2. If $k \geq 2$, then

   $$\p{1 + \frac{x}{k}}^{-k} \leq \p{1 + \frac{x}{2}}^{-2} \in L^1\p{\pco{0, \infty}}.$$

   Thus, because $\abs{\cos\frac{x}{k}} \leq 1$, we may apply the dominated convergence theorem:

   $$\begin{aligned} \lim_{k\to\infty} \int_0^\infty \p{1 + \frac{x}{k}}^{-k} \cos\p{\frac{x}{k}} \,\diff{x} &= \int_0^\infty \lim_{k\to\infty} \p{1 + \frac{x}{k}}^{-k} \cos\p{\frac{x}{k}} \,\diff{x} \\ &= \int_0^\infty e^{-x} \cos\p{0} \,\diff{x} \\ &= \int_0^\infty e^{-x} \,\diff{x} \\ &= 1. \end{aligned}$$