題目 4. (10%) Evaluate the limit. limn→∞1n(n+1)(n+3)(n+5)⋯(3n−1)n\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+3)(n+5)\cdots(3n-1)}n→∞limn1n(n+1)(n+3)(n+5)⋯(3n−1) 解答 解法一 取對數，這是 Riemann 和： ln[1n∏k=0n−1(n+2k+1)n]→∫01ln(1+2t) dt=3ln3−22.\ln\left[\frac1n\sqrt[n]{\prod_{k=0}^{n-1}(n+2k+1)}\right] \to\int_0^1\ln(1+2t)\,\mathrm{d}t =\frac{3\ln3-2}{2}.lnn1nk=0∏n−1(n+2k+1)→∫01ln(1+2t)dt=23ln3−2. 故極限為 33e.\boxed{\frac{3\sqrt3}{e}}.e33.
