问题及解答

证明: 对任意 $n\geqslant 1$, 有

\[\sum_{k=n+1}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}.\]

证明: 

\[
\begin{split}
\mathrm{RHS}&=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}\\
&=(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{2n-1}+\frac{1}{2n})-2(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n})\\
&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k}\\
&=\sum_{k=n+1}^{2n}\frac{1}{k}\\
&=\mathrm{LHS}.
\end{split}
\]

证毕.