回復 1# larson 的帖子
令 \( f(x) = \frac{1}{x} - \frac{1}{[x]+1} \), \( x>0 \)。則 \( f(x) \) 恆正。
\( \begin{aligned}\int_{1}^{n}f(x)dx & =\int_{1}^{n}\frac{1}{x}dx-\int_{1}^{n}\frac{1}{[x]+1}dx\\
& =\log n-\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\right)
\end{aligned} \)
由 \( f(x) \) 恆正知上式遞增
而 \( f(x) = \frac{[x]+1-x}{x([x]+1)}\leq\frac{1}{x^{2}} \)，又 \( \int_1^\infty \frac{1}{x^2} dx <\infty \)
由比較判別法知 \( \int_1^\infty f(x)dx \) 收斂，故 \( \log n-\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\right) \) 收斂 (或遞增有上界得收斂)
因此原數列亦收斂
