Ta có; \(B=1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2022}+\frac{1}{2023}\)
\(=1+\frac12+\frac13+\cdots+\frac{1}{2023}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)
\(=1+\frac12+\ldots+\frac{1}{2023}-1-\frac12-\cdots-\frac{1}{1011}=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2023}\)
=C
=>B-C=0