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Question Number 180899 by alcohol last updated on 19/Nov/22

$$H_n=\underset{k=1}{\sum ^n}\frac{1}{k}$$$$showthat{H}_{2n}-H_n=\underset{k=1}{\sum ^n}(\frac{1}{2k-1}-\frac{1}{2k})$$

Answered by Frix last updated on 19/Nov/22

$$H_{2n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n}=$$$$=\frac{1}{1}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n}=$$$$=\underset{k=1}{\sum ^n}\frac{1}{2k-1}+\underset{k=1}{\sum ^n}\frac{1}{2k}=\underset{k=1}{\sum ^n}\frac{1}{2k-1}+\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{k}$$$$\Rightarrow$$$$H_{2n}-H_n=\underset{k=1}{\sum ^n}\frac{1}{2k-1}+\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{k}-\underset{k=1}{\sum ^n}\frac{1}{k}=$$$$=\underset{k=1}{\sum ^n}\frac{1}{2k-1}-\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{k}=\underset{k=1}{\sum ^n}\frac{1}{2k-1}-\underset{k=1}{\sum ^n}\frac{1}{2k}=$$$$=\underset{k=1}{\sum ^n}(\frac{1}{2k-1}-\frac{1}{2k})$$

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