底下這樣寫不知道會不會有問題，歡迎批評指教。
16.
\( \displaystyle \lim_{n \to \infty }\frac{1}{n\sqrt{n}}\sum_{k=1}^{n}\sqrt{\frac{2k-1}{2}} \)
\( \displaystyle = \lim_{n \to \infty }\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{2k-1}{2n}} \)
\( \displaystyle \leq  \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{2k}{2n}}\)
\( \displaystyle =\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}} \)
\( \displaystyle = \int_{0}^{1}\sqrt{x}dx\)
\( \displaystyle =\frac{2}{3} \)
又
\( \displaystyle \lim_{n \to \infty }\frac{1}{n\sqrt{n}}\sum_{k=1}^{n}\sqrt{\frac{2k-1}{2}} \)
\( \displaystyle = \lim_{n \to \infty }\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{2k-1}{2n}} \)
\( \displaystyle = \lim_{n \to \infty }\frac{1}{n}\sum_{k=2}^{n}\sqrt{\frac{2k-1}{2n}} \)
\( \displaystyle = \lim_{n \to \infty }\frac{1}{n}\sum_{k=1}^{n-1}\sqrt{\frac{2k+1}{2n}} \)
\( \displaystyle = \lim_{n \to \infty }\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{2k+1}{2n}} \)
\( \displaystyle \geq  \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{2k}{2n}}\)
\( \displaystyle =\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}} \)
\( \displaystyle = \int_{0}^{1}\sqrt{x}dx\)
\( \displaystyle =\frac{2}{3} \)
所以， \( \displaystyle \lim_{n \to \infty }\frac{1}{n\sqrt{n}}\sum_{k=1}^{n}\sqrt{\frac{2k-1}{2}}=\frac{2}{3} \)

對齁~那剛好是中點~這樣一行就結束了~感謝老王老師~你太帥了~