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Question Number 180828 by Vynho last updated on 17/Nov/22

$$H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$$$H_{2n}=?compute{H}_{2n}-H_{n}and{H}_{n+1}-H_n$$

Commented by Frix last updated on 17/Nov/22

$$\mathrm{obviously}{H}_{n+1}-H_n=\frac{1}{n+1}$$$$\mathrm{I}\mathrm{believe}\mathrm{that}\underset{n\to \mathrm{\infty }}{\lim }(H_{kn}-H_n)=\ln k\mathrm{for}k\in \mathbb{N}$$

Answered by Frix last updated on 19/Nov/22

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

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